Question

Calculate the change in pH when 49.0 mL of a 0.700 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

use:

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

= 4.745+ log {1/1}

= 4.745

Lets calculate the final pH

mol of NaOH added = 0.7M *0.049 L = 0.0343 mol

CH3COOH will react with OH- to form CH3COO-

Before Reaction:

mol of CH3COO- = 1.0 M *1.0 L

mol of CH3COO- = 1 mol

mol of CH3COOH = 1.0 M *1.0 L

mol of CH3COOH = 1 mol

after reaction,

mol of CH3COO- = mol present initially + mol added

mol of CH3COO- = (1 + 0.0343) mol

mol of CH3COO- = 1.0343 mol

mol of CH3COOH = mol present initially - mol added

mol of CH3COOH = (1 - 0.0343) mol

mol of CH3COOH = 0.9657 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 {1.034/0.9657}

= 4.775

pH change = |final pH - initial pH|

= | 4.775 - 4.745 |

= 0.03

Answer: 0.03