Question:

Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?

Answer:

1. Pure liquids and solids can be ignored while writing the equilibrium constant expression because they are not affected by changes in temperature or pressure. The equilibrium constant expression is only used to describe the equilibrium state of a reaction when the temperature and pressure remain constant. Therefore, pure liquids and solids do not need to be included in the expression.

Question:

What is meant by the conjugate acid-base pair? Find the conjugate acid/base for the following species. HNO₂,CN^-,HClO₄F^-,OH^-,CO₃^2- and S^2-

Answer:

What is meant by the conjugate acid-base pair?

A conjugate acid-base pair is a pair of compounds or ions that are related by the transfer of a single proton (H⁺). The acid is the species that donates the proton, and the base is the species that accepts the proton.

Find the conjugate acid/base for the following species.

HNO₂: HNO₂ (acid) and NO₂^- (base) CN^-: HCN (acid) and CN^- (base) HClO₄F^-: HClO₄F^- (acid) and ClO₄F^- (base) OH^-: H₂O (acid) and OH^- (base) CO₃^2-: HCO₃^- (acid) and CO₃2- (base) S^2-: H₂S (acid) and S2- (base)

Question:

Which of the followings are Lewis acids? H₂O,BF3,H⁺, and NH₄⁺

Answer:

1. H₂O: Not a Lewis acid

2. BF3: Lewis acid

3. H⁺: Lewis acid

4. NH₄⁺: Lewis acid

Question:

Write the conjugate acids for the following Bronsted bases NH₂−,NH₃ and HCOO^-.

Answer:

NH₂−: H₃O⁺ NH₃: NH₄⁺ HCOO^-: HCOOH

Question:

A sample of pure PCl₅ was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl₅ was found to be 0.5×10−1mol L−1. If value of Kc is 8.3×10^-3, what are the concentrations of PCl₃ and Cl₂ at equilibrium? PCl₅(g)⇌PCl₃(g)+Cl₂(g)

Answer:

1. Write down the equilibrium reaction: PCl₅(g)⇌PCl₃(g)+Cl₂(g)

2. Calculate the initial concentration of PCl₅: [PCl₅] = 0.5 × 10^-1 mol L-1

3. Calculate the equilibrium concentrations of PCl₃ and Cl₂ using the equilibrium expression: Kc = [PCl₃][Cl₂]/[PCl₅] [PCl₃] = [Cl₂] = (Kc × [PCl₅])1/2 [PCl₃] = [Cl₂] = (8.3 × 10-3 × 0.5 × 10^-1)1/2 [PCl₃] = [Cl₂] = 0.0291 mol L-1

Question:

Predict which of the following reaction will have appreciable concentration of reactants and products: a) Cl₂(g)⇌2Cl(g);Kc=5×10^(−39) b) Cl₂(g)+2NO(g)⇌2NOCl(g);Kc=3.7×10^8 c) Cl₂(g)+2NO₂(g)⇌2NO₂Cl(g);Kc=1.8

Answer:

a) Cl₂(g)⇌2Cl(g);Kc=5×10^(−39) This reaction will not have appreciable concentrations of reactants and products because the Kc value is too small.

b) Cl₂(g)+2NO(g)⇌2NOCl(g);Kc=3.7×10^8 This reaction will have appreciable concentrations of reactants and products because the Kc value is relatively large.

c) Cl₂(g)+2NO₂(g)⇌2NO₂Cl(g);Kc=1.8 This reaction will have appreciable concentrations of reactants and products because the Kc value is relatively large.

Question:

The ionization constant of HF,HCOOH and HCN at 298K are 6.8×10^(−4),1.8×10^(−4) and 4.8×10⁽⁻⁹⁾ respectively. Calculate the ioniation constants of the corresponding conjugate base.

Answer:

1. The conjugate base of HF is F⁻, the conjugate base of HCOOH is COO⁻ and the conjugate base of HCN is CN⁻.

2. The ionization constant of F⁻ is 6.8×10⁽⁻⁴⁾.

3. The ionization constant of COO⁻ is 1.8×10⁽⁻⁴⁾.

4. The ionization constant of CN⁻ is 4.8×10^(-9).

Question:

Classify the following species into Lewis acids and Lewis bases and show how these act as Lewis acid/base: (a) OH^- (b) F^- (c) H⁺ (d) BCl3.

Answer:

(a) OH^- - Lewis Base: Acts as a Lewis base by donating a lone pair of electrons.

(b) F^- - Lewis Base: Acts as a Lewis base by donating a lone pair of electrons.

(c) H⁺ - Lewis Acid: Acts as a Lewis acid by accepting a lone pair of electrons.

(d) BCl3 - Lewis Acid: Acts as a Lewis acid by accepting a lone pair of electrons.

Question:

The degree of ionization of a 0.1 M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the pKa of bromoacetic acid.

Answer:

1. Calculate the concentration of the bromoacetic acid molecules: 0.1 M x (1 - 0.132) = 0.088 M

2. Calculate the concentration of the bromoacetic acid ions: 0.1 M x 0.132 = 0.0132 M

3. Calculate the hydrogen ion concentration: 0.0132 M

4. Calculate the pH of the solution: -log(0.0132) = 1.89

5. Calculate the pKa of bromoacetic acid: pKa = -log(Ka) = -log(0.0132) = 2.88

Question:

Calculate the pH of the resultant mixtures: a) 10 mL of 0.2M Ca(OH)₂ + 25 mL of 0.1M HCl b) 10 mL of 0.01M H₂SO4+ 10 mL of 0.01M Ca(OH)₂ c) 10 mL of 0.1M H₂SO4 + 10 mL of 0.1M KOH

Answer:

a) 10 mL of 0.2M Ca(OH)₂ + 25 mL of 0.1M HCl

Step 1: Calculate the moles of each compound.

Moles of Ca(OH)₂ = (0.2 M)(0.01 L) = 0.002 mol

Moles of HCl = (0.1 M)(0.025 L) = 0.0025 mol

Step 2: Calculate the amount of H⁺ ions produced.

H⁺ ions from Ca(OH)₂ = 0.002 mol

H⁺ ions from HCl = 0.0025 mol

Step 3: Calculate the total amount of H⁺ ions.

Total H⁺ ions = 0.002 mol + 0.0025 mol = 0.0045 mol

Step 4: Calculate the pH of the solution.

pH = -log(0.0045) = 2.35

b) 10 mL of 0.01M H₂SO4+ 10 mL of 0.01M Ca(OH)₂

Step 1: Calculate the moles of each compound.

Moles of H₂SO4 = (0.01 M)(0.01 L) = 0.0001 mol

Moles of Ca(OH)₂ = (0.01 M)(0.01 L) = 0.0001 mol

Step 2: Calculate the amount of H⁺ ions produced.

H⁺ ions from H₂SO4 = 0.0001 mol

H⁺ ions from Ca(OH)₂ = 0.0001 mol

Step 3: Calculate the total amount of H⁺ ions.

Total H⁺ ions = 0.0001 mol + 0.0001 mol = 0.0002 mol

Step 4: Calculate the pH of the solution.

pH = -log(0.0002) = 3.7

c) 10 mL of 0.1M H₂SO4 + 10 mL of 0.1M KOH

Step 1: Calculate the moles of each compound.

Moles of H₂SO4 = (0.1 M)(0.01 L) = 0.001 mol

Moles of KOH = (0.1 M)(0.01 L) = 0.001 mol

Step 2: Calculate the amount of H⁺ ions produced.

H⁺ ions from H₂SO4 = 0.001 mol

H⁺ ions from KOH = 0.001 mol

Step 3: Calculate the total amount of H⁺ ions.

Total H⁺ ions = 0.001 mol + 0.001 mol = 0.002 mol

Step 4: Calculate the pH of the solution.

pH = -log(0.002) = 2.7

Question:

Write the expression for the equilibrium constant, Kc for each of the following reactions. (i) 2NOCl(g)⇌2NO(g)+Cl₂(g) (ii) 2Cu(NO₃)2(s)⇌2CuO(s)+4NO₂(g)+O₂(g) (iii) CH₃COOC₂H₅(aq)+H₂O(1)⇌CH₃COOH(aq)+C₂H₅OH(aq) (iv) Fe3+(aq)+3OH^-(aq)⇌Fe(OH)3(s) (v) I₂(s)+5F2⇌2IF5

Answer:

(i) Kc = [NO][Cl₂]²/[NOCl]²

(ii) Kc = [CuO][NO₂]⁴/[Cu(NO₃)2]

(iii) Kc = [CH₃COOH][C₂H₅OH]/[CH₃COOC₂H₅][H₂O]

(iv) Kc = [Fe(OH)3]/[Fe3+][OH^-]^3

(v) Kc = [IF5]²/[I₂][F2]⁵

Question:

At a certain temperature and total pressure of 10⁵ Pa, iodine vapour contains 40% by volume of I atoms. I₂(g)⇌2I(g) Calculate Kp for the equilibrium.

Answer:

Step 1: Identify the equilibrium equation: I₂(g)⇌2I(g)

Step 2: Calculate the mole fractions of I₂ and I: X(I₂) = 0.4 X(I) = 0.8

Step 3: Calculate the partial pressures of I₂ and I: P(I₂) = 0.4 x 10⁵ Pa = 4 x 10⁴ Pa P(I) = 0.8 x 10⁵ Pa = 8 x 10⁴ Pa

Step 4: Calculate Kp for the equilibrium: Kp = (P(I)²) / P(I₂) = (8 x 10⁴)² / (4 x 10⁴) = 64

Question:

Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression.

Answer:

Answer:

1. Pure liquids and solids can be ignored while writing the equilibrium constant expression because they are not involved in the chemical reaction and remain unchanged throughout the reaction.

2. The equilibrium constant expression only applies to substances that are in the gas or aqueous phase, because these are the only substances that are affected by the reaction and can change in concentration.

3. Therefore, since pure liquids and solids are not affected by the reaction, they can be omitted from the equilibrium constant expression.

Question:

Nitric oxide reacts with bromine and gives nitrosyl-bromide as per reaction given below: 2NO(g)+Br₂(g)⇌2NOBr(g) When 0.087 mole of NO and 0.0437 mole of Br₂ are mixed is a closed container at constant temperature, 0.0518 mole of NOBr is obtained at equilibrium. Calculate the equilibrium amount of nitric oxide and bromine.

Answer:

Step 1: Write down the given information:

NO (initial): 0.087 moles Br₂ (initial): 0.0437 moles NOBr (equilibrium): 0.0518 moles

Step 2: Calculate the equilibrium amount of nitric oxide:

NO (equilibrium) = 0.087 - 0.0518 = 0.0352 moles

Step 3: Calculate the equilibrium amount of bromine:

Br₂ (equilibrium) = 0.0437 - 0.0518 = -0.008 moles

Therefore, the equilibrium amount of nitric oxide is 0.0352 moles and the equilibrium amount of bromine is -0.008 moles.

Question:

A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI(g) is 0.04 atm. What of KP for the given equilibrium? 2HI(g)⇌H₂(g)+I₂(g) : A : 6 B : 16 C : 4 D : 2

Answer:

Answer: C

Step 1: Determine the equilibrium equation: 2HI(g)⇌H₂(g)+I₂(g)

Step 2: Calculate the equilibrium constant (Kp): Kp = (PH₂ * PI₂) / (PHI)²

Step 3: Substitute the given values: Kp = (0.04 * 0.04) / (0.2)²

Step 4: Calculate the equilibrium constant: Kp = 4

Step 5: Select the correct answer: Answer: C

Question:

The equilibrium constant expression for a gas reaction is, KC=([NH₃]⁴[O₂]⁵)/([NO]⁴[H₂O]^6) The balanced chemical equation corresponding to this expression is. A : 4NO(g)+6H₂O(g)⇌4NH₃(g)+5O₂(g) B : 4NH₃(g)+5O₂(g)⇌4NO(g)+6H₂O(g) C : Both of above D : None of these

Answer:

Answer: C: Both of above

Question:

The ester, ethyl acetate is formed by the reaction between ethanol and acetic acid and their equilibrium is represented as: CH₃COOH(l)+C₂H₅OH(l)⇌CH₃COOC₂H₅(aq)+H₂O(l) (a) Write the concentration ratio (reaction quotient), Qe, for this reaction. Note that water is not in excess and is not a solvent in this reaction. (b) At 293 K, if one starts with 1.00 mole of acetic acid and 0.180 mole of ethanol, there is 0.171 mole of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant. (c) Starting with 0.500 mole of ethanol and 1.000 mole of acetic acid and maintaining it at 293 K, 0.214 mole of ethyl acetate is found after some time. Has equilibrium been reached?

Answer:

a) Qe = [CH₃COOC₂H₅]² / [CH₃COOH][C₂H₅OH]

b) Kc = [CH₃COOC₂H₅]² / [CH₃COOH][C₂H₅OH] = 0.171² / (1.00)(0.180) = 2.18

c) No, equilibrium has not been reached.

Question:

Calculate: a) ΔG⁰. b) the equilibrium constant for the formation of NO₂ from NO and O₂ at 298 K. NO(g)+21O₂(g)⇌NO₂(g)
where ΔfG⁰(NO₂)=52.0 kJ/mol ΔfG⁰(NO)=87.0 kJ/mol ΔfG⁰(O₂)=0 kJ/mol.

Answer:

a) ΔG⁰ = ΔfG⁰(NO₂) - (ΔfG⁰(NO) + ΔfG⁰(O₂)) = 52.0 kJ/mol - (87.0 kJ/mol + 0 kJ/mol) = -35.0 kJ/mol

b) The equilibrium constant for the formation of NO₂ from NO and O₂ at 298 K can be calculated using the equation: K = e^(-ΔG⁰/RT) = e^(-(-35.0 kJ/mol)/(8.314 J/mol K × 298 K)) = 0.544

Question:

Which of the following reactions will get affected by increasing the pressure? Also, mention whether change will cause the reaction to go into forward or backward direction. (i) COCl₂(g)⇌CO(g)+Cl₂(g) (ii) CH₄(g)+2S2(g)⇌CS2(g)+2H₂S(g) (iii) CO₂(g)+C(s)⇌2CO(g) (iv) 2H₂(g)+CO(g)⇌CH₃OH(g) (v) CaCO₃(s)⇌CaO(s)+CO₂(g) (vi) 4NH₃(g)+5O₂(g)⇌4NO(g)+6H₂O(g)

Answer:

(i) Increasing the pressure will cause the reaction to go into the forward direction.

(ii) Increasing the pressure will cause the reaction to go into the forward direction.

(iii) Increasing the pressure will cause the reaction to go into the backward direction.

(iv) Increasing the pressure will cause the reaction to go into the backward direction.

(v) Increasing the pressure will cause the reaction to go into the backward direction.

(vi) Increasing the pressure will cause the reaction to go into the forward direction.

Question:

Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction: CH₄(g)+H₂O(g)⇌CO(g)+3H₂(g) (a) Write as expression for Kp for the above reaction. (b) How will the values of Kp and composition of equilibrium mixture be affected by (i) increasing the pressure, (ii) Increasing the temperature and (iii) Using a catalyst?

Answer:

(a) Kp = [CO(g)][3H₂(g)]/[CH₄(g)][H₂O(g)]

(b) (i) Increasing the pressure will increase the value of Kp and shift the equilibrium to the right.

(ii) Increasing the temperature will increase the value of Kp and shift the equilibrium to the right.

(iii) Using a catalyst may increase the rate of reaction but will not affect the value of Kp or the composition of the equilibrium mixture.

Question:

Describe the effect of following components on the equilibrium of the given reaction. (a) addition of H₂ (b) addition of CH₃OH (c) removal of CO (d) removal of CH₃OH 2H₂(g)+CO(g)⇌CH₃OH(g)

Answer:

(a) The addition of H₂ will shift the equilibrium to the right, meaning that more CH₃OH will be produced.

(b) The addition of CH₃OH will shift the equilibrium to the left, meaning that less CH₃OH will be produced.

(c) The removal of CO will shift the equilibrium to the left, meaning that less CH₃OH will be produced.

(d) The removal of CH₃OH will shift the equilibrium to the right, meaning that more CH₃OH will be produced.

Question:

At 473 K, equilibrium constant Kc for decomposition of phosphorus pentachloride, PCl₅ is 8.3×10⁽⁻³⁾. If decomposition is depicted as, PCl₅(g)⇌PCl₃(g)+Cl₂(g) ΔrH⊖=124.0kJmol^(-1) (a) Write an expression for Kc for the reaction. (b) What is the value of Kc for the reverse reaction at the same temperature? (c) What would be the effect on Kc if (i) more PCl₅ is added (ii) pressure is increased (iii) the temperature is increased ?

Answer:

(a) Kc = [PCl₃][Cl₂]/[PCl₅]

(b) Kc for the reverse reaction at the same temperature is 8.3 x 10⁽⁻³⁾.

(c) (i) If more PCl₅ is added, Kc will remain unchanged. (ii) If pressure is increased, Kc will increase. (iii) If the temperature is increased, Kc will increase.

Question:

The value of Kc for the reaction 3O₂(g)⇌2O₃(g) is 2.0×10⁽⁻⁵⁰⁾ at 25°C. If the equilibrium concentration of O₂ in air at 25oC is 1.6×10⁽⁻²⁾ M, what is the concentration of O₃?

Answer:

1. Kc = 2.0x10^(-50)
2. Equilibrium concentration of O₂ = 1.6x10⁽⁻²⁾ M
3. Use the equation Kc = [O₃]²/[O₂]^3
4. [O₃]²/[O₂]^3 = 2.0x10^(-50)
5. [O₂]^3 = 2.0x10^(-50) x [O₃]²
6. [O₂]^3 = 2.0x10^(-50) x [O₃]²
7. [O₂] = (2.0x10^(-50) x [O₃]²)^(1/3)
8. [O₂] = (2.0x10^(-50) x [O₃]²)^(1/3)
9. [O₂] = (2.0x10^(-50) x 1.6x10⁽⁻²⁾²)^(1/3)
10. [O₂] = (3.2x10^(-53))^(1/3)
11. [O₂] = 1.6x10^(-17) M
12. Insert [O₂] into the equation Kc = [O₃]²/[O₂]^3
13. [O₃]²/1.6x10^(-17)^3 = 2.0x10^(-50)
14. [O₃]² = 2.0x10^(-50) x 1.6x10^(-17)^3
15. [O₃]² = 6.4x10^(-84)
16. [O₃] = (6.4x10^(-84))^(1/2)
17. [O₃] = 2.5x10^(-42) M

Question:

The ionization constant of phenol is 1.0×10^(-10). What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization if the solution is also 0.01 M in sodium phenolate?

Answer:

1. The concentration of phenolate ion in a 0.05 M solution of phenol is 1.0×10⁽⁻¹⁰⁾ M.

2. The degree of ionization of the solution is calculated by the formula: Degree of ionization = [Phenolate ion]/[Phenol] = 1.0×10⁽⁻¹⁰⁾ M / (0.05 M + 0.01 M) = 0.0196

Question:

The ionization constant of acetic acid is 1.74×10⁽⁻⁵⁾. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the solution and its pH.

Answer:

1. The degree of dissociation of acetic acid (α) can be calculated using the equation: α = [H⁺]² / (K_a * [HA])

2. Substituting the values, we get: α = [H⁺]² / (1.74×10^(-5) * 0.05)

3. Since the concentration of acetic acid in the solution is 0.05 M, the concentration of H⁺ ion is equal to the degree of dissociation (α): [H⁺] = α

4. Substituting the value of [H⁺] in the equation, we get: α = α² / (1.74×10^(-5) * 0.05)

5. Solving for α, we get: α = 0.0164

6. The concentration of acetate ion in the solution is equal to the degree of dissociation (α) multiplied by the concentration of acetic acid (0.05 M): [acetate ion] = α * 0.05 = 0.00820 M

7. The pH of the solution can be calculated using the equation: pH = -log[H⁺]

8. Substituting the value of [H⁺], we get: pH = -log(0.0164) = 1.79

Question:

It has been found that the pH of a 0.01 M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its pKa.

Answer:

1. Calculate the concentration of the anion: The concentration of the anion is equal to the concentration of the acid, which is 0.01 M.

2. Calculate the ionization constant of the acid: The ionization constant of the acid is equal to 10^-4.15, which is equal to 0.00012.

3. Calculate the pKa of the acid: The pKa of the acid is equal to -log(0.00012), which is equal to 4.92.

Question:

If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pH ?

Answer:

Step 1: Calculate the molar mass of KOH. Molar mass of KOH = 56.10 g/mol

Step 2: Calculate the moles of KOH. Moles of KOH = 0.561 g/56.10 g/mol = 0.01000 moles

Step 3: Calculate the concentration of potassium ions. Concentration of potassium ions = 0.01000 moles/200 mL = 0.05000 M

Step 4: Calculate the concentration of hydrogen ions. Concentration of hydrogen ions = 0.01000 moles/200 mL = 0.05000 M

Step 5: Calculate the concentration of hydroxyl ions. Concentration of hydroxyl ions = 0.01000 moles/200 mL = 0.05000 M

Step 6: Calculate the pH. pH = -log[0.05000] = 1.30

Question:

The pH of 0.1 M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.

Answer:

Step 1: Calculate the ionization constant of the acid.

The ionization constant of cyanic acid (HCNO) is Ka = [H⁺][CNO-]/[HCNO]

Step 2: Calculate the concentration of H⁺.

The pH of the solution is given as 2.34, so [H⁺] = 10^-2.34 = 4.3 x 10^-3 mol/L

Step 3: Calculate the concentration of CNO-.

The degree of ionization is given as α = [CNO-]/[HCNO], so [CNO-] = α[HCNO]

Step 4: Calculate the degree of ionization.

The degree of ionization is given by α = [CNO-]/[HCNO] = 4.3 x 10^-3/0.1 = 0.043

Step 5: Substitute the values into the equation for Ka.

Ka = [H⁺][CNO-]/[HCNO] = (4.3 x 10^-3)(0.043)/0.1 = 1.8 x 10^-4

Question:

A 0.02 M solution of pyridinium hydrochloride has pH=3.44. Calculate the ionization constant of pyridine.

Answer:

Step 1: Calculate the concentration of the pyridinium cation (C+) in the solution.

C+ = 0.02 M

Step 2: Calculate the concentration of the pyridinium anion (A-) in the solution.

A- = 0.02 M

Step 3: Calculate the ionization constant of pyridine, Kp.

Kp = [C+][A-]/[Pyridine] = (0.02 M)(0.02 M)/(0.02 M) = 0.04

Question:

The ionization constant of chloroacetic acid is 1.35×10⁽⁻³⁾. What will be the pH of 0.1 M acid and its 0.1 M sodium salt solution?

Answer:

Step 1: Calculate the concentration of hydrogen ions in 0.1 M chloroacetic acid solution.

The ionization constant of chloroacetic acid is given as 1.35×10⁽⁻³⁾.

The concentration of hydrogen ions can be calculated using the equation:

[H⁺] = K × [acid]

[H⁺] = 1.35 × 10⁽⁻³⁾ × 0.1 = 1.35 × 10⁽⁻⁴⁾

Step 2: Calculate the pH of 0.1 M chloroacetic acid solution.

The pH of the solution can be calculated using the equation:

pH = -log[H⁺]

pH = -log(1.35 × 10⁽⁻⁴⁾) = 3.87

Step 3: Calculate the concentration of hydrogen ions in 0.1 M sodium salt solution.

The concentration of hydrogen ions in the sodium salt solution can be calculated using the equation:

[H⁺] = [salt]

[H⁺] = 0.1

Step 4: Calculate the pH of 0.1 M sodium salt solution.

The pH of the solution can be calculated using the equation:

pH = -log[H⁺]

pH = -log(0.1) = 1

Question:

The ionization constant of benzoic acid is 6.46×10⁽⁻⁵⁾ and Ksp for silver benzoate is 2.5×10⁽⁻¹³⁾. How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water?

Answer:

1. Calculate the molar solubility of silver benzoate in pure water by using the Ksp expression:

Ksp = [Ag+][C7H₅O₂^-]

[Ag+] = 2.5 x 10^(-13) / [C7H₅O₂^-]

[Ag+] = 2.5 x 10^(-13) / 6.46 x 10^(-5)

[Ag+] = 0.003868 M

2. Calculate the molar solubility of silver benzoate in a buffer of pH 3.19 by using the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

pKa + log([A-]/[HA]) = 3.19

log([A-]/[HA]) = 3.19 - 6.46

log([A-]/[HA]) = -3.27

[A-]/[HA] = 0.04

Using the ionization constant of benzoic acid, calculate the molar solubility of silver benzoate:

Ksp = [Ag+][C7H₅O₂^-]

[Ag+] = 2.5 x 10^(-13) / [C7H₅O₂^-]

[Ag+] = 2.5 x 10^(-13) / (0.04 x 6.46 x 10^(-5))

[Ag+] = 0.0099 M

3. Calculate the ratio of solubility of silver benzoate in a buffer of pH 3.19 compared to its solubility in pure water:

Ratio = (0.0099 M) / (0.003868 M)

Ratio = 2.56

Therefore, silver benzoate is 2.56 times more soluble in a buffer of pH 3.19 compared to its solubility in pure water.

Question:

The concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen sulphide is 1.0×10^(−19)M. If 10 mL of this is added to 5 mL of 0.04 M solution of FeSO4, MnCl₂, ZnCl₂ and CdCl₂. In which of these solutions precipitation will take place?

Answer:

1. Calculate the total concentration of sulphide ion in the given solution: The total concentration of sulphide ion in the given solution is (0.1 M) + (1.0×10^(−19)M) = 0.1 M

2. Calculate the total volume of the solution: The total volume of the solution is (10 mL) + (5 mL) = 15 mL

3. Calculate the sulphide ion concentration in the solution: The sulphide ion concentration in the solution is (0.1 M) / (15 mL) = 6.67 × 10^(-20)M

4. Determine which of the solutions will precipitate: The FeSO4, MnCl₂ and ZnCl₂ solutions will not precipitate because the sulphide ion concentration is too low. However, the CdCl₂ solution will precipitate because the sulphide ion concentration is higher than the solubility product of CdCl₂.

Question:

At 700 K, the equilibrium constant for the reaction; H₂(g)+I₂(g)⇌2HI(g) is 54.8. If 0.5 mol litre^-1 of HI(g) is present at equilibrium at 700 K, what are the concentrations of H₂(g) and I₂(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700 K.

Answer:

1. Write the equilibrium expression: K = [HI]² / [H₂][I₂]

2. Substitute the given values: K = 54.8 = (0.5)² / [H₂][I₂]

3. Solve for [H₂] and [I₂]: [H₂] = (0.5)² / (54.8 * 0.5) = 0.009 mol litre^-1 [I₂] = (0.5)² / (54.8 * 0.5) = 0.009 mol litre^-1

Question:

KP=0.04atm at 899K for the equilibrium shown below. What is the equilibrium concentration of C₂H6 when it is placed in a flask at 4.0atm pressure and allowed to come to equilibrium? C₂H6(g)⇌C₂H6(g)+H₂(g) : A : 7.24atm B : 3.62atm C : 1atm D : 1.5atm

Answer:

Answer: Step 1: Calculate the equilibrium constant (KP) for the given reaction: KP = (Pressure of C₂H6) / (Pressure of H₂)

Step 2: Calculate the equilibrium concentration of C₂H6 when it is placed in a flask at 4.0atm pressure and allowed to come to equilibrium: KP = (Pressure of C₂H6) / (Pressure of H₂) = 4.0atm / 1atm = 4.0.

Step 3: Calculate the equilibrium concentration of C₂H6: C₂H6 = 4.0 x KP = 4.0 x 0.04atm = 0.16atm

Question:

Equilibrium constant, KC for the reaction at 500K is 0.061. N₂(g)+3H₂(g)⇌2NH₃(g). At a particular time, the analysis shows that composition of the reaction mixture is 3.0 molL^(-1) N₂, 2.0 molL^(-1) H₂ and 0.5 molL^(-1) NH₃. Is the reaction at equilibrium and if not in which direction does the reaction tend to proceed to reach equilibrium?

Answer:

1. Calculate the reaction quotient, Q: Q = (3.0 molL^(-1))(2.0 molL^(-1))^3/(0.5 molL^(-1))² = 24

2. Compare Q with Kc: 24 > 0.061

3. Since Q > Kc, the reaction is not at equilibrium and tends to proceed in the reverse direction to reach equilibrium.

Question:

At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO₂ in equilibrium with solid carbon has 90.55% of CO by mass. C(s)+CO₂(g)⇌2CO(g) Calculate Kc for the reaction at the above temperature.

Answer:

Step 1: Write out the chemical equation for the reaction. C(s) + CO₂(g) ⇌ 2CO(g)

Step 2: Calculate the moles of CO and CO₂ in the reaction. Moles of CO = 90.55%/100 x 1 mole = 0.9055 moles Moles of CO₂ = (1 - 90.55%)/100 x 1 mole = 0.0945 moles

Step 3: Calculate the equilibrium concentrations of CO and CO₂. Equilibrium concentration of CO = 0.9055 moles/1 liter = 0.9055 M Equilibrium concentration of CO₂ = 0.0945 moles/1 liter = 0.0945 M

Step 4: Calculate Kc for the reaction. Kc = ([CO]2/[CO₂])/[C(s)] Kc = (0.9055)2/0.0945)/1 Kc = 8.5

Question:

The equilibrium constant for the following reactions is 1.6×105 at 1024 K. H₂(g)+Br₂(g)⇌2HBr(g) Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024 K.

Answer:

1. Write the equilibrium expression: K = [HBr]² / [H₂][Br₂]

2. Substitute the equilibrium constant: K = (10.0 bar)² / [H₂][Br₂]

3. Solve for [H₂] and [Br₂]: [H₂] = [Br₂] = (10.0 bar) / sqrt(1.6 x 10⁵)

4. Calculate the equilibrium pressure of all gases: Equilibrium pressure of H₂ = Br₂ = 0.5 bar Equilibrium pressure of HBr = 10.0 bar

Question:

Dihydrogen gas used in Habers process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of CO and H₂. In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction, CO(g)+H₂O(g)⇌CO₂(g)+H₂(g) If a reaction vessel at 400oC is charged with an equimolar mixture of CO and steam such that PCO=PH₂O=4.0bar, what will be the partial pressure of H₂ at equilibrium? KP=0.1 at 400oC?

Answer:

Step 1: Identify the given values in the question. Given values:

• Dihydrogen gas used in Habers process is produced by reacting methane from natural gas with high temperature steam
• first stage of two stage reaction involves the formation of CO and H₂
• second stage involves the water gas shift reaction
• reaction vessel at 400oC is charged with an equimolar mixture of CO and steam such that PCO=PH₂O=4.0bar
• KP=0.1 at 400oC

Step 2: Calculate the equilibrium partial pressures of CO₂ and H₂ using the water gas shift reaction equation. Equilibrium partial pressures of CO₂ and H₂: PCO₂ = 4.0bar PH₂ = 0.1*4.0bar = 0.4bar

Step 3: The partial pressure of H₂ at equilibrium is 0.4bar.

Question:

What is Kc for the following equilibrium when the equilibrium concentration of each substance is [SO₂]=0.60M,[O₂]=0.82M,[SO₃]=1.90M2SO₂⁺O₂⇌2SO₃.

Answer:

1. Write the equilibrium expression: Kc = [SO₃]2 / [SO₂][O₂]2

2. Substitute the equilibrium concentrations into the expression: Kc = (1.90M)2 / (0.60M)(0.82M)2

3. Solve for Kc: Kc = 6.84

Question:

Find out the value of KC for each of the following equilibrium respectively from the value of KP. (i) 2NOCl(g)⇌2NO(g)+Cl₂(g);KP=1.8×10⁽⁻²⁾ at 500 K (ii) CaCO₃(s)⇌CaO(s)+CO₂(g);KP=167 at 1073 K

Answer:

(i) Kc = KP x (RT)^(Δn)

Kc = 1.8 × 10⁽⁻²⁾ x (8.314 JK^-1mol^-1 x 500K)^(2)

Kc = 2.67 × 10⁽⁻⁴⁾

(ii) Kc = KP x (RT)^(Δn)

Kc = 167 x (8.314 JK^-1mol^-1 x 1073K)⁽¹⁾

Kc = 1.53 × 10^(3)

Question:

What will be the conjugate bases for the Bronsted acids HF,H₂SO4 and HCO₃−?

Answer:

Answer:

1. For HF, the conjugate base will be F^-.

2. For H₂SO4, the conjugate bases will be SO₄^2- and H₂O.

3. For HCO₃−, the conjugate base will be CO₃^2-.

Question:

The species: H₂O,HCO₃^-,HSO₄^- and NH₃ can act both as Bronsted acids and bases. For each case give the corresponding conjugate acid and base.

Answer:

H₂O: Acid: H₃O⁺ Base: OH^-

HCO₃^-: Acid: H₂CO₃ Base: HCO₃^-

HSO4-: Acid: H₂SO4 Base: SO4–

NH₃: Acid: NH₄⁺ Base: NH₂^-

Question:

The first ionization constant of H₂S is 9.1×10⁽⁻⁸⁾. Calculate the concentration of HS^- ion in its 0.1 M solution. How will this concentration be affected if the solution is 0.1M in HCl also? If the second dissociation constant of H₂S is 1.2×10⁽⁻¹³⁾, calculate the concentration of S^2- under both conditions.

Answer:

The first ionization constant of H₂S is 9.1×10⁽⁻⁸⁾.

Calculate the concentration of HS^- ion in its 0.1 M solution:

The concentration of HS^- ion in a 0.1 M solution of H₂S is 0.1 x 9.1×10⁽⁻⁸⁾ = 9.1×10⁽⁻⁹⁾ M.

How will this concentration be affected if the solution is 0.1M in HCl also?

If the solution is 0.1M in HCl, the concentration of HS^- ion will remain the same. The HCl will not affect the concentration of HS^- ion.

If the second dissociation constant of H₂S is 1.2×10⁽⁻¹³⁾, calculate the concentration of S^2- under both conditions:

The concentration of S^2- in a 0.1 M solution of H₂S is 0.1 x 1.2×10⁽⁻¹³⁾ = 1.2×10⁽⁻¹⁴⁾ M.

The concentration of S^2- in a 0.1M solution of H₂S and HCl will be the same as in a 0.1M solution of H₂S only. The HCl will not affect the concentration of S^2-.

Question:

Calculate the degree of ionization of 0.05 M acetic acid if its PKa value is 4.74. How is the degree of dissociation affected when its solution also contains (a) 0.01M (b) 0.1M in HCl ?

Answer:

Answer:

Step 1: Calculate the degree of ionization of 0.05 M acetic acid if its PKa value is 4.74.

The degree of ionization of 0.05 M acetic acid is 0.0017.

Step 2: How is the degree of dissociation affected when its solution also contains (a) 0.01M (b) 0.1M in HCl ?

(a) When the solution also contains 0.01M HCl, the degree of dissociation is increased because HCl is a strong acid and will be completely ionized in the solution, thus increasing the total concentration of ions in the solution.

(b) When the solution also contains 0.1M HCl, the degree of dissociation is significantly increased because HCl is a strong acid and will be completely ionized in the solution, thus significantly increasing the total concentration of ions in the solution.

Question:

The solubility of Sr(OH)₂ at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.

Answer:

1. Calculate the molar solubility of Sr(OH)₂: 19.23 g/L x (1 mol/119.98 g) = 0.159 mol/L

2. Calculate the concentration of strontium ions: 0.159 mol/L x (1 mol Sr₂⁺/1 mol Sr(OH)₂) = 0.159 mol/L

3. Calculate the concentration of hydroxyl ions: 0.159 mol/L x (2 mol OH^-/1 mol Sr(OH)₂) = 0.318 mol/L

4. Calculate the pH of the solution: -log[OH^-] = -log(0.318) = 0.505

Question:

The ionization constant of propanoic acid is 1.32×10⁽⁻⁵⁾. Calculate the degree of ionization of the acid in its 0.05 M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl ?

Answer:

1. The ionization constant of propanoic acid is 1.32 x 10^(-5).

2. Calculate the degree of ionization of the acid in its 0.05 M solution: The degree of ionization can be calculated using the equation: Degree of ionization = [H⁺]/[HA] where [H⁺] is the concentration of the acid and [HA] is the concentration of the acid before ionization.

In this case, [H⁺] = 0.05 M and [HA] = 0.05 M, so the degree of ionization is 1.

3. Calculate the pH of the solution: The pH of the solution can be calculated using the equation: pH = -log[H⁺]

In this case, [H⁺] = 0.05 M, so the pH of the solution is 1.3.

4. Calculate the degree of ionization of the acid if the solution is 0.01M in HCl: The degree of ionization can be calculated using the equation: Degree of ionization = [H⁺]/[HA]

In this case, [H⁺] = 0.01 M and [HA] = 0.01 M, so the degree of ionization is 1.

Question:

Ionic product of water at 310 K is 2.7×10⁽⁻¹⁴⁾. What is the pH of neutral water at this temperature?

Answer:

Step 1: Calculate the concentration of hydronium ions, [H₃O⁺], in neutral water at 310 K.

Step 2: Use the equation [H₃O⁺] = K_w/[H₂O] to calculate the concentration of water at 310 K.

Step 3: Calculate the pH of neutral water at 310 K by using the equation pH = -log[H₃O⁺].

Question:

What is the maximum concentration of an equimolar solution of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is not precipitation of iron sulphide? (For FeS, Ksp=6.3×10^(-18))

Answer:

1. Calculate the molar concentration of ferrous sulphate and sodium sulphide separately.
2. Ksp = [Fe2+]*[S2-]²
3. Substitute the molar concentration of ferrous sulphate and sodium sulphide in the Ksp equation.
4. Rearrange the equation to solve for [S2-].
5. Substitute the value of [S2-] in the molar concentration of sodium sulphide to calculate the maximum concentration of sodium sulphide.
6. The maximum concentration of an equimolar solution of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is not precipitation of iron sulphide will be the maximum concentration of sodium sulphide calculated in step 5.

Question:

Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298 K from their solubility product constants. Determine also the molarities of individual ions.

Answer:

1. Find the solubility product constants (Ksp) for silver chromate, barium chromate, ferric hydroxide, lead chloride, and mercurous iodide.

2. Calculate the solubilities of each compound at 298 K using the Ksp values.

3. Calculate the molarities of individual ions by dividing the solubility of each compound by its respective formula weight.

Question:

The solubility product constants are: Silver chromate, 1.1×10^(-12) Barium chromate, 1.2×10^(-10) Ferric hydroxide, 1.0×10^(−38) Lead dichloride, 1.6×10⁽⁻⁵⁾ Mercurous iodide, 4.5×10^(−29)

Answer:

1. Silver chromate: The solubility product constant for silver chromate is 1.1×10^(-12).

2. Barium chromate: The solubility product constant for barium chromate is 1.2×10^(-10).

3. Ferric hydroxide: The solubility product constant for ferric hydroxide is 1.0×10^(−38).

4. Lead dichloride: The solubility product constant for lead dichloride is 1.6×10⁽⁻⁵⁾.

5. Mercurous iodide: The solubility product constant for mercurous iodide is 4.5×10^(−29).

Question:

What is the maximum concentration of equimolar solutions of ferrous sulphateand sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide? (For iron sulphide, Ksp=6.3×10^(-18)).

Answer:

1. Calculate the Ksp of iron sulphide (Ksp=6.3×10^(-18)).

2. Determine the molar concentrations of ferrous sulphate and sodium sulphide. The molar concentrations must be equal for there to be no precipitation of iron sulphide.

3. Calculate the product of the molar concentrations of ferrous sulphate and sodium sulphide. This must equal the Ksp of iron sulphide (6.3×10^(-18)).

4. Solve for the maximum concentration of the equimolar solutions of ferrous sulphate and sodium sulphide. The maximum concentration is equal to the Ksp of iron sulphide (6.3×10^(-18)).

Question:

What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M? 2ICl(g)⇌I₂(g)+Cl₂(g);Kc=0.14

Answer:

Step 1: Write the equilibrium equation: 2ICl(g)⇌I₂(g)+Cl₂(g);Kc=0.14

Step 2: Calculate the equilibrium concentrations of each substance:

[I₂] = 0.78 M / 2 = 0.39 M

[Cl₂] = 0.78 M / 2 = 0.39 M

[ICl] = 0.78 M - 0.39 M - 0.39 M = 0.00 M

Question:

A sample of pure PCl₅ was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl₅ was found to be 0.5×10^(-1)mol L^(-1). Pcl5(g)⇌PCl₃(g)+Cl₂(g),KC=8.3×10⁽⁻³⁾ mol L^(-1)

Answer:

1. Calculate the partial pressures of PCl₃ and Cl₂ at equilibrium: PCl₃ = 0.5 × 10^(-1) mol L^(-1) × (8.3 × 10⁽⁻³⁾ mol L^(-1)) / (1 - 0.5 × 10^(-1) mol L^(-1)) = 0.415 mol L^(-1) Cl₂ = 0.5 × 10^(-1) mol L^(-1) × (8.3 × 10⁽⁻³⁾ mol L^(-1)) / (1 - 0.5 × 10^(-1) mol L^(-1)) = 0.415 mol L^(-1)

2. Calculate the mole fraction of PCl₃ and Cl₂ at equilibrium: PCl₃ = 0.415 mol L^(-1) / (0.415 + 0.415) = 0.5 Cl₂ = 0.415 mol L^(-1) / (0.415 + 0.415) = 0.5

Question:

One of the reaction that takes place in producing steel from iron ore is the reduction of iron (II) oxide by carbon monoxide to give iron metal and CO₂ . FeO(s)+CO(g)⇌Fe(s)+CO₂(g);Kp=0.265 atm at 105 K What are the equilibrium partial pressures of CO and CO₂ at 1050 K if the initial pressure are PCO=1.4atm and PCO₂=0.80atm?

Answer:

1. Calculate the reaction quotient, Q: Q = (PCO₂/PCO)² = (0.8/1.4)² = 0.357

2. Compare Q with Kp: Q < Kp, so the reaction is not at equilibrium.

3. Calculate the equilibrium partial pressures of CO and CO₂ using the equation: PCO₂/PCO = (Kp)^1/2 = (0.265)^1/2 = 0.514

Therefore, the equilibrium partial pressures of CO and CO₂ at 1050 K are: PCO = 1.4atm and PCO₂ = 0.714atm

Question:

Does the number of moles of reaction products increase, decrease or remain same when each of the following equiibria is subjected to a decrease in pressure bp increasing the volume? i) PCl₅⇌PCl₃(g)+Cl₂(g) ii) CaO(s)+CO₂(g)⇌CaCO₃(s) iii) 3Fe(s)+4H₂O(g)⇌Fe3O4(s)+4H₂(g)

Answer:

i) Decrease ii) Decrease iii) Increase

Question:

The reaction, CO(g)+3H₂(g)⇌CH₄(g)+H₂O(g), is at equilibrium at 1300 K in a 1L flask. It also contain 0.30 mol of CO, 0.10 mol of H₂ and 0.02 mol of H₂O and an unknown amount of CH₄ in the flask. Determine the concentration of CH₄ in the mixture. The equilibrium constant, Kc for the reaction at the given temperature is 3.90.

Answer:

1. Calculate the moles of CH₄ present in the flask:

Moles of CH₄ = 0.30 mol of CO + 0.10 mol of H₂ - 0.02 mol of H₂O = 0.38 mol of CH₄

2. Calculate the concentration of CH₄ in the mixture:

Concentration of CH₄ = 0.38 mol of CH₄/1L = 0.38 M

Question:

Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below:

Answer:

pH = 7

Hydrogen ion concentration = 10^-7 moles/L

Question:

(a) Human muscle-fluid, 6.83 (b) Human stomach fluid, 1.2 (c) Human blood, 7.38 (d) Human saliva, 6.4.

Answer:

a) pH of Human muscle-fluid = 6.83

b) pH of Human stomach fluid = 1.2

c) pH of Human blood = 7.38

d) pH of Human saliva = 6.4

Question:

The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8,5.0,4.2,2.2 and 7.8 respectively. Calculate corresponding hydrogen ion concentration in each.

Answer:

Milk: Hydrogen ion concentration = 10^-6.8 = 1.25 x 10^-7 moles/L

Black Coffee: Hydrogen ion concentration = 10^-5.0 = 1.00 x 10^-5 moles/L

Tomato Juice: Hydrogen ion concentration = 10^-4.2 = 6.31 x 10^-5 moles/L

Lemon Juice: Hydrogen ion concentration = 10^-2.2 = 1.58 x 10^-3 moles/L

Egg White: Hydrogen ion concentration = 10^-7.8 = 3.98 x 10^-8 moles/L

Question:

The ionization constant of nitrous acid is 4.5×10−4. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.

Answer:

Step 1: Calculate the concentration of nitrous acid, HNO₂.

Nitrous acid, HNO₂, is formed by the hydrolysis of sodium nitrite, NaNO₂. The degree of hydrolysis is equal to the ratio of the concentration of nitrous acid to the concentration of sodium nitrite.

Degree of hydrolysis = [HNO₂]/[NaNO₂]

[HNO₂] = Degree of hydrolysis x [NaNO₂]

[HNO₂] = 0.04 x 4.5 x 10-4

[HNO₂] = 1.8 x 10-5 M

Step 2: Calculate the pH of the solution.

The pH of a solution is determined by the concentration of hydrogen ions, [H⁺]. The ionization constant of nitrous acid, Ka, is 4.5 x 10-4.

Ka = [H⁺][NO₂^-]/[HNO₂]

[H⁺] = Ka x [HNO₂]/[NO₂^-]

[H⁺] = 4.5 x 10-4 x 1.8 x 10-5/0.04

[H⁺] = 8.1 x 10-8 M

pH = -log[H⁺]

pH = 7.09

Question:

What is the minimum volume of water required to dissolve 1 g of calcium sulphate at 298 K? (For calcium sulphate, Ksp=9.1×10^(-6)).

Answer:

Step 1: Calculate the molar mass of calcium sulphate. Molar mass of calcium sulphate = (40 g + 32 g + 64 g) = 136 g/mol

Step 2: Calculate the molar concentration of calcium sulphate. Molar concentration of calcium sulphate = 1 g/136 g/mol = 0.00735 mol/L

Step 3: Calculate the solubility product (Ksp) of calcium sulphate. Ksp = 9.1 × 10^(-6)

Step 4: Calculate the concentration of calcium and sulphate ions. Concentration of calcium ion = sqrt(Ksp/[sulphate]) = sqrt(9.1 × 10^(-6)/[0.00735]) = 0.085 mol/L

Concentration of sulphate ion = 0.00735 mol/L

Step 5: Calculate the volume of water required to dissolve 1 g of calcium sulphate. Volume of water = (1 g/136 g/mol) × (1 L/0.085 mol/L) = 15.88 L

Question:

Write the conjugate acids for the following Bronsted bases : NH₂−,NH₃ and HCOO^-.

Answer:

NH₂−: H₃O⁺ NH₃: NH₄⁺ HCOO^-: HCOOH

Question:

The concentration of hydrogen ion in a sample of soft drink is 3.8×10⁽⁻³⁾M. What is its pH value?

Answer:

Step 1: Calculate the hydrogen ion concentration (H⁺) in the soft drink.

H⁺ = 3.8 x 10⁽⁻³⁾ M

Step 2: Calculate the pH value of the soft drink.

pH = -log(H⁺)

pH = -log(3.8 x 10⁽⁻³⁾)

pH = 2.42

Question:

Assuming complete dissociation, calculate the pH of the following solutions: (a) 0.003 M HCl (b) 0.005 M NaOH (c) 0.002 M HBr (d) 0.002 M KOH

Answer:

(a) 0.003 M HCl

Step 1: Calculate the molarity of H⁺ ions:

Molarity of H⁺ ions = 0.003 M

Step 2: Calculate the pH of the solution:

pH = -log[H⁺]

pH = -log(0.003)

pH = 2.52

Question:

For the following equilibrium, KC=6.3×10¹⁴ at 1000K. NO(g)+O₃(g)⇌NO₂(g)+O₂(g). Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is KC, for the reverse reaction? A : 15.9×10⁽⁻¹⁵⁾ B : 1.59×10⁽⁻¹⁵⁾ C : 5×10⁽⁻¹⁵⁾ D : 9×10⁽⁻¹⁵⁾

Answer:

Answer: B : 1.59×10⁽⁻¹⁵⁾

Explanation:

The equilibrium constant for the reverse reaction can be calculated using the equilibrium constant for the forward reaction and the equilibrium constant expression:

K_c = (K_c)_forward / (K_c)_reverse

Therefore, (K_c)_reverse = (K_c)_forward / K_c

Given that (K_c)_forward = 6.3×10¹⁴ and K_c = 6.3×10¹⁴, (K_c)_reverse = 6.3×10¹⁴ / 6.3×10¹⁴ = 1.59×10⁽⁻¹⁵⁾.

Therefore, the correct answer is B : 1.59×10⁽⁻¹⁵⁾.

Question:

Reaction between Nitrogen and Oxygen takes place as follows:- 2N₂(g)+O₂(g)⇌2N₂O(g) If a mixture of 0.482 mol of N₂ and 0.933 mol of O₂ is placed in a reaction vessel of volume 10L and allowed to form N₂O at a temperature for which Ke=2.0×10^(−37), determine the composition of the equilibrium mixture.

Answer:

1. Write the balanced equation for the reaction: 2N₂(g) + O₂(g) ⇌ 2N₂O(g)

2. Calculate the initial moles of reactants: N₂: 0.482 mol O₂: 0.933 mol

3. Calculate the equilibrium moles of each species: N₂: 0.482 mol O₂: 0.933 mol N₂O: 0.966 mol

4. Calculate the equilibrium concentrations of each species: N₂: 0.0482 mol/L O₂: 0.0933 mol/L N₂O: 0.0966 mol/L

5. Calculate the equilibrium partial pressures of each species: N₂: 0.482 atm O₂: 0.933 atm N₂O: 0.966 atm

Question:

A mixture of 1.57 mol of N₂,1.92 mol of H₂ and 8.13 mol of NH₃ is introduced in a 20L reaction vessel at 500K.At this temperature the equilibrium constant, Kc for the reaction N₂(g)+3H₂(g)⇌2NH₃(g) is 1.7×10². Is the reaction mixture at equilibrium? If not what is the direction of net reaction?

Answer:

1. Calculate the molar concentration of each component at 500K: N₂: 1.57 mol/20L = 0.0785 mol/L H₂: 1.92 mol/20L = 0.096 mol/L NH₃: 8.13 mol/20L = 0.4065 mol/L

2. Calculate the equilibrium constant, Kc: Kc = [NH₃]²/[N₂][H₂]^3 Kc = (0.4065)² / (0.0785)(0.096)^3 Kc = 1.7 × 10²

3. Compare the calculated Kc to the given Kc: Given Kc = 1.7 × 10² Calculated Kc = 1.7 × 10² Since the calculated Kc is equal to the given Kc, the reaction mixture is at equilibrium.

Therefore, the direction of the net reaction is none.

Question:

One mole of H₂O and one mole of CO are taken in 10L vessel and heated to 725 K. At equilibrium 40% of water (by mass) reacts with CO according to the equation, H₂O(g)+CO(g)⇌H₂(g)+CO₂(g) What is the value of equilibrium constant (Kc) for the reaction. A : 44 B : 4.4 C : 0.44 D : 2.22

Answer:

Answer: B. 4.4

Step 1: Write the balanced chemical equation for the reaction. H₂O(g) + CO(g) ⇌ H₂(g) + CO₂(g)

Step 2: Calculate the moles of each reactant and product. Moles of H₂O = 1 mol Moles of CO = 1 mol Moles of H₂ = 0.4 mol Moles of CO₂ = 0.6 mol

Step 3: Calculate the equilibrium constant (Kc). Kc = (H₂)(CO₂)/(H₂O)(CO) Kc = (0.4 mol)(0.6 mol)/(1 mol)(1 mol) = 0.24

Step 4: Calculate the value of Kc in terms of the given options. A : 44 => 44 x 0.24 = 10.56 (incorrect) B : 4.4 => 4.4 x 0.24 = 1.056 (correct) C : 0.44 => 0.44 x 0.24 = 0.1056 (incorrect) D : 2.22 => 2.22 x 0.24 = 0.528 (incorrect)

Therefore, the correct answer is B. 4.4

Question:

The ionization constant of dimethylamine is 5.4×10^(−4). Calculate its degree of ionization in its 0.02 M solution. What percentage of dimethylamine is ionized if the solution is also 0.1M in NaOH?

Answer:

1. Calculate the degree of ionization:

Degree of ionization = [Me2NH⁺]/[Me2NH]

[Me2NH⁺] = 5.4 × 10⁽⁻⁴⁾ × 0.02M = 1.08 × 10⁽⁻⁴⁾

[Me2NH] = 0.02M - 1.08 × 10⁽⁻⁴⁾ = 0.019992M

Degree of ionization = 1.08 × 10⁽⁻⁴⁾ / 0.019992M = 0.05402

2. Calculate the percentage of dimethylamine that is ionized:

Percentage of dimethylamine that is ionized = (Degree of ionization × 100) / (0.1M NaOH)

Percentage of dimethylamine that is ionized = (0.05402 × 100) / (0.1M NaOH) = 54.02%

Question:

The solubility product constant of Ag₂CrO₄ and AgBr is 1.1×10^(-12) and 5.0×10⁽⁻¹³⁾ respectively. Calculate the ratio of the molarities of their saturated solutions.

Answer:

1. Write down the solubility product constants for Ag₂CrO₄ and AgBr:

Ag₂CrO₄: Ksp = 1.1×10^(-12) AgBr: Ksp = 5.0×10⁽⁻¹³⁾

2. Calculate the molar solubility of Ag₂CrO₄:

Molar solubility of Ag₂CrO₄ = (Ksp)^1/2 Molar solubility of Ag₂CrO₄ = (1.1×10^(-12))^1/2 Molar solubility of Ag₂CrO₄ = 3.3×10^(-6)

3. Calculate the molar solubility of AgBr:

Molar solubility of AgBr = (Ksp)^1/2 Molar solubility of AgBr = (5.0×10⁽⁻¹³⁾)^1/2 Molar solubility of AgBr = 7.1×10^(-7)

4. Calculate the ratio of the molarities of their saturated solutions:

Ratio of molarities = (Molar solubility of Ag₂CrO₄)/(Molar solubility of AgBr) Ratio of molarities = 3.3×10^(-6)/7.1×10^(-7) Ratio of molarities = 4.6

Question:

Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate, Ksp=7.4×10⁽⁻⁸⁾ ).

Answer:

1. Calculate the molar concentration of iodate ions (IO₃^-) in the solution after mixing.

The molar concentration of iodate ions in the solution after mixing is 0.002 M.

2. Calculate the molar concentration of chlorate ions (ClO₃^-) in the solution after mixing.

The molar concentration of chlorate ions in the solution after mixing is also 0.002 M.

3. Calculate the molar concentration of cupric ions (Cu₂⁺) in the solution after mixing.

The molar concentration of cupric ions in the solution after mixing is 0.004 M.

4. Calculate the molar concentration of copper iodate (Cu(IO₃)2) in the solution after mixing.

The molar concentration of copper iodate in the solution after mixing is 0.

5. Compare the molar concentration of copper iodate to the Ksp of copper iodate.

The molar concentration of copper iodate is 0, which is lower than the Ksp of copper iodate (7.4 x 10^(-8)). Therefore, it will not lead to precipitation of copper iodate.

Question:

The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ions in it.

Answer:

Step 1: Convert the pH of 3.76 to a negative logarithm.

Step 2: The negative logarithm of 3.76 is -0.42.

Step 3: Calculate the concentration of hydrogen ions by taking 10 to the power of -0.42.

Step 4: The concentration of hydrogen ions in the vinegar is 0.40 M.

Question:

Calculate the pH of the following solutions: a) 2 g of TlOH dissolved in water to give 2 litre of solution. b) 0.3 g of Ca(OH)₂ dissolved in water to give 500 mL of solution. c) 0.3 g of NaOH dissolved in water to give 200 mL of solution. d) 1 mL of 13.6 M HCl is diluted with water to give 1 litre of solution.

Answer:

a) To find the pH of the solution, we first need to calculate the molarity of the solution. The molar mass of TlOH is 203.9 g/mol, so the molarity of the solution is:

Molarity = (2 g TlOH)/(203.9 g/mol) x (2 L solution)/(1000 mL/L) Molarity = 0.0098 mol/L

The pH of the solution can then be calculated using the equation:

pH = -log[H⁺]

[H⁺] = 10^-pH

[H⁺] = 10^-14/0.0098 [H⁺] = 1.02 x 10^-12

pH = -log(1.02 x 10^-12) pH = 11.99

b) To find the pH of the solution, we first need to calculate the molarity of the solution. The molar mass of Ca(OH)₂ is 74.1 g/mol, so the molarity of the solution is:

Molarity = (0.3 g Ca(OH)₂)/(74.1 g/mol) x (500 mL solution)/(1000 mL/L) Molarity = 0.00402 mol/L

The pH of the solution can then be calculated using the equation:

pH = -log[H⁺]

[H⁺] = 10^-pH

[H⁺] = 10^-14/0.00402 [H⁺] = 2.48 x 10^-12

pH = -log(2.48 x 10^-12) pH = 11.60

c) To find the pH of the solution, we first need to calculate the molarity of the solution. The molar mass of NaOH is 40 g/mol, so the molarity of the solution is:

Molarity = (0.3 g NaOH)/(40 g/mol) x (200 mL solution)/(1000 mL/L) Molarity = 0.0075 mol/L

The pH of the solution can then be calculated using the equation:

pH = -log[H⁺]

[H⁺] = 10^-pH

[H⁺] = 10^-14/0.0075 [H⁺] = 1.33 x 10^-12

pH = -log(1.33 x 10^-12) pH = 11.88

d) To find the pH of the solution, we first need to calculate the molarity of the solution. The molarity of the initial solution is 13.6 M, so the molarity of the diluted solution is:

Molarity = (1 mL HCl)/(1000 mL solution) x (13.6 M HCl) Molarity = 0.0136 M

The pH of the solution can then be calculated using the equation:

pH = -log[H⁺]

[H⁺] = 10^-pH

[H⁺] = 10^-14/0.0136 [H⁺] = 7.35 x 10^-13

pH = -log(7.35 x 10^-13) pH = 12.13