Answer to Question #122114 in Physical Chemistry for vandana jian

Question

(a) Calculate the pH of a solution resulting from adding 4 mL of 0.2 M of NaOH solution to
16 mL of a buffer solution containing 0.1 M CH3COOH ( Ka = 1.8X10-5 ) and 0.2 M
CH3COONa ?
(b) Calculate the pH of a solution resulting from adding 6 mL of 0.2 M HCl to 14 mL of a buffer
solution containing 0.2 M NH3 ( Kb = 1.75X10-5 ) and 0.1 M NH4Cl

Answer 1

Answer.

(a) (first part)

Concentration of NaOH after addition to 16 ml Buffer solution = S_NaOH (M) (Let)

Total volume of solution after addition of 4 ml 0.2 (M) NaOH to 16 ml Buffer solution = (16+4) = 20 ml

According to the formula,

V₁ x S₁ = V₂ x S₂

20 ml x S_NaOH = 4 ml x 0.2 (M)

S_NaOH = 0.04 (M)

Total volume of solution after addition of 4 ml NaOH to 16 ml Buffer solution = (16+4) = 20 ml

Concentration of CH₃CO₂H considering the dilution effect = S_CH3CO2H (M) (Let)

According to the formula,

V₁ x S₁ = V₂ x S₂

20 ml x S_CH3CO2H = 16 ml x 0.1 (M)

S_CH3CO2H = 0.08 (M)

Concentration of CH₃CO₂Na considering the dilution effect = S _CH3CO2Na (M) (Let)

According to the formula,

V₁ x S₁ = V₂ x S₂

20 ml x S_CH3CO2Na = 16 ml x 0.2 (M)

S_CH3CO2Na = 0.16 (M)

According to Henderson’ s equation for acidic buffer,

pH = pKa + log ([salt]/[acid])

Here after addition of NaOH, the OH^-  ion will react with the acetate to produce acetic acid. So concentration of acid will increase and salt will decrease.

Thus, now, concentration of acetic acid will be = (0.08 + 0.04) (M) = 0.12 (M)

Thus, now, concentration of acetate salt will be = (0.16 - 0.04) (M) = 0.12 (M)

Thus applying Henderson’s equation,

pH = pKa + log(0.12/0.12)

pH = pKa = -log(1.8 x 10^-5) = 4.74 (Ans)

(b) (second part)

Concentration of HCl after addition to 14 ml Buffer solution = S_HCl (M) (Let)

Total volume of solution after addition of 6 ml 0.2 (M) HCl to 14 ml Buffer solution = (14+6) = 20 ml

According to the formula,

V₁ x S₁ = V₂ x S₂

20 ml x S_HCl = 6ml x 0.2 (M)

S_HCl = 0.06 (M)

Total volume of solution after addition of 6 ml 0.2 (M) HCl to 14 ml Buffer solution = (14+6) = 20 ml

Concentration of NH₃ considering the dilution effect = S_NH3 (M) (Let)

According to the formula,

V₁ x S₁ = V₂ x S₂

20 ml x S _NH3 = 14 ml x 0.2 (M)

S _NH3 = 0.14 (M)

Concentration of NH₄Cl considering the dilution effect = S_NH4Cl (M) (Let)

According to the formula,

V₁ x S₁ = V₂ x S₂

20 ml x S_NH4Cl = 14 ml x 0.1 (M)

S_NH4Cl = 0.07 (M)

According to Henderson’ s equation for basic buffer,

pOH = pKb + log ([salt]/[base])

Here after addition of HCl, the H⁺ ion will react with the ammonia to produce ammonium ion. So concentration of base will decrease and salt will increase.

Thus, now, concentration of ammonia will be = (0.14 - 0.06) (M) = 0.08 (M)

Thus, now, concentration of ammonium chloride salt will be = (0.07 + 0.06) (M) = 0.13 (M)

Thus applying Henderson’s equation,

pOH = pKb + log (0.13/0.08)

pOH = -log(1.75 x 10^-5) + 0.210

pOH = 4.96

pH = (14-4.96) = 9.04 (Ans.)