Question

A 300.0 mL buffer solution is 0.220 M in acetic acid and 0.220 M in sodium acetate.
What is the initial pH of this solution?
What is the pH after addition of 0.0100 mol of HCl?
What is the pH after addition of 0.0100 mol of NaOH?

Please explain I don't understand how to do this. thank you!

Answer 1

1)

Ka = 1.8*10^-5

pKa = - log (Ka)

= - log(1.8*10^-5)

= 4.745

use:

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

= 4.745+ log {0.22/0.22}

= 4.745

Answer: 4.74

2)

mol of HCl added = 0.01 mol

CH3COO- will react with H+ to form CH3COOH

Before Reaction:

mol of CH3COO- = 0.22 M *0.3 L

mol of CH3COO- = 0.066 mol

mol of CH3COOH = 0.22 M *0.3 L

mol of CH3COOH = 0.066 mol

after reaction,

mol of CH3COO- = mol present initially - mol added

mol of CH3COO- = (0.066 - 0.01) mol

mol of CH3COO- = 0.056 mol

mol of CH3COOH = mol present initially + mol added

mol of CH3COOH = (0.066 + 0.01) mol

mol of CH3COOH = 0.076 mol

Ka = 1.8*10^-5

pKa = - log (Ka)

= - log(1.8*10^-5)

= 4.745

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

use:

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

= 4.745+ log {5.6*10^-2/7.6*10^-2}

= 4.612

Answer: 4.61

3)

mol of NaOH added = 0.01 mol

CH3COOH will react with OH- to form CH3COO-

Before Reaction:

mol of CH3COO- = 0.22 M *0.3 L

mol of CH3COO- = 0.066 mol

mol of CH3COOH = 0.22 M *0.3 L

mol of CH3COOH = 0.066 mol

after reaction,

mol of CH3COO- = mol present initially + mol added

mol of CH3COO- = (0.066 + 0.01) mol

mol of CH3COO- = 0.076 mol

mol of CH3COOH = mol present initially - mol added

mol of CH3COOH = (0.066 - 0.01) mol

mol of CH3COOH = 0.056 mol

Ka = 1.8*10^-5

pKa = - log (Ka)

= - log(1.8*10^-5)

= 4.745

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

use:

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

= 4.745+ log {7.6*10^-2/5.6*10^-2}

= 4.877

Answer: 4.88