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Calculate the pH of 0.379 L of a 0.25 M acetic acid - 0.34 M sodium...

Calculate the pH of 0.379 L of a 0.25 M acetic acid - 0.34 M sodium acetate buffer before and after the addition of 0.0058 mol of KOH. Assume that the volume remains constant. (Ka of acitic acid is 1.8⋅10−5)

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Answer #1

No of mole of CH3COO- =(0.34mol/1000ml)×379ml= 0.12886

No of mole of CH3COOH=(0.25mol/1000ml)×379ml= 0.09475

KOH react with CH3COOH

OH-+ CH3COOH -------> CH3COO-

0.0058 mol KOH react with 0.0058 CH3COOH

remaining moles of CH3COOH = 0.09475 - 0.0058 = 0.08895

moles of CH3COO- after addn = 0.12886 + 0.0058 = 0.13466

[ CH3COOH ] = (0.08895/379)×1000=0.2347M

[ CH3COO- ] = (0.13466/379)×1000 = 0.3553M

Henderson equation is

pH = pKa + log ( [ A- ]/[HA])

= 4.74 + log (0.3553/0.2347)

= 4.74 + 0.18

= 4.92

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