Question

Part A

Calculate the pH of 0.376 L of a 0.18 M acetic acid - 0.35 M sodium acetate buffer before and after the addition of 0.0065 mol of KOH. Assume that the volume remains constant. (Ka of acitic acid is 1.8⋅10−5) Express your answer using three significant figures.

Part B

Calculate the pH of 0.376 L of a 0.18 M acetic acid - 0.35 M sodium acetate buffer before and after the addition of 0.0065 mol of HBr. Assume that the volume remains constant. (Ka of acetic acid is 1.8⋅10−5)

Express your answer using three significant figures.

Answer 1

A)

Lets calculate the initial pH

Ka = 1.8*10^-5

pKa = - log (Ka)

= - log(1.8*10^-5)

= 4.7447

use:

pH = pKa + log {[conjugate base]/[acid]}

= 4.7447+ log {0.35/0.18}

= 5.03

Answer: 5.03

Lets calculate the final pH

mol of KOH added = 0.0065 mol

CH3COOH will react with OH- to form CH3COO-

Before Reaction:

mol of CH3COO- = 0.35 M *0.376 L

mol of CH3COO- = 0.1316 mol

mol of CH3COOH = 0.18 M *0.376 L

mol of CH3COOH = 0.0677 mol

after reaction,

mol of CH3COO- = mol present initially + mol added

mol of CH3COO- = (0.1316 + 0.0065) mol

mol of CH3COO- = 0.1381 mol

mol of CH3COOH = mol present initially - mol added

mol of CH3COOH = (0.0677 - 0.0065) mol

mol of CH3COOH = 0.0612 mol

since volume is both in numerator and denominator, we can use mol instead of concentration

use:

pH = pKa + log {[conjugate base]/[acid]}

= 4.7447+ log {0.1381/0.0612}

= 5.0983

pH is 5.10

Answer: 5.10

B)

Lets calculate the initial pH

use:

pH = pKa + log {[conjugate base]/[acid]}

= 4.7447+ log {0.35/0.18}

= 5.03

Answer: 5.03

Lets calculate the final pH

mol of HBr added = 0.0065 mol

CH3COO- will react with H+ to form CH3COOH

Before Reaction:

mol of CH3COO- = 0.35 M *0.376 L

mol of CH3COO- = 0.1316 mol

mol of CH3COOH = 0.18 M *0.376 L

mol of CH3COOH = 0.0677 mol

after reaction,

mol of CH3COO- = mol present initially - mol added

mol of CH3COO- = (0.1316 - 0.0065) mol

mol of CH3COO- = 0.1251 mol

mol of CH3COOH = mol present initially + mol added

mol of CH3COOH = (0.0677 + 0.0065) mol

mol of CH3COOH = 0.0742 mol

since volume is both in numerator and denominator, we can use mol instead of concentration

use:

pH = pKa + log {[conjugate base]/[acid]}

= 4.7447+ log {0.1251/0.0742}

= 4.9717

pH is 4.97

Answer: 4.97