Question

Calculate the change in pH when 55.0 mL of a 0.620 M solution of NaOH is added to 1.00 L of a solution that is 1.00 M in sodium acetate and 1.00 M in acetic acid.

Answer 1

Lets calculate the initial pH

Ka = 1.8*10^-5

pKa = - log (Ka)

= - log(1.8*10^-5)

= 4.745

we have below equation to be used:

This is Henderson–Hasselbalch equation

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

= 4.745+ log {1/1}

= 4.745

Lets calculate the final pH

mol of NaOH added = 0.62M *55.0 mL = 34.1 mmol

CH3COOH will react with OH- to form CH3COO-

Before Reaction:

mol of CH3COO- = 1.0 M *1000.0 mL

mol of CH3COO- = 1000 mmol

mol of CH3COOH = 1.0 M *1000.0 mL

mol of CH3COOH = 1000 mmol

after reaction,

mol of CH3COO- = mol present initially + mol added

mol of CH3COO- = (1000 + 34.1) mmol

mol of CH3COO- = 1034.1 mmol

mol of CH3COOH = mol present initially - mol added

mol of CH3COOH = (1000 - 34.1) mmol

mol of CH3COOH = 965.9 mmol

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

we have below equation to be used:

This is Henderson–Hasselbalch equation

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

= 4.745+ log {1.034*10^3/9.659*10^2}

= 4.774

pH change = |final pH - initial pH|

= | 4.774 - 4.745 |

= 0.0296

Answer: 0.0296