Question

A buffer solution is 0.78 M in acetic acid and 0.22 M in sodium acetate. Calculate the solution pH after adding 0.80 g of solid NaOH to 100.0 mL of the buffer solution. Ka of acetic acid is 1.8  10−5 . Assume negligible volume change.

Answer 1

Molar mass of NaOH,

MM = 1*MM(Na) + 1*MM(O) + 1*MM(H)

= 1*22.99 + 1*16.0 + 1*1.008

= 39.998 g/mol

mass(NaOH)= 0.80 g

use:

number of mol of NaOH,

n = mass of NaOH/molar mass of NaOH

=(0.8 g)/(40 g/mol)

= 2*10^-2 mol

mol of NaOH added = 20.0 mmol

CH3COOH will react with OH- to form CH3COO-

Before Reaction:

mol of CH3COO- = 0.22 M *100.0 mL

mol of CH3COO- = 22 mmol

mol of CH3COOH = 0.78 M *100.0 mL

mol of CH3COOH = 78 mmol

after reaction,

mol of CH3COO- = mol present initially + mol added

mol of CH3COO- = (22 + 20.0) mmol

mol of CH3COO- = 42.001 mmol

mol of CH3COOH = mol present initially - mol added

mol of CH3COOH = (78 - 20.0) mmol

mol of CH3COOH = 57.999 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

use:

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

= 4.745+ log {42/58}

= 4.605

Answer: 4.60