Question

A buffer is prepared by mixing 160.0mL of 0.15M formic acid, a monoprotic acid, and 140.0mL of 0.25M sodium formate. Calculate the pH. Now using 100.0mL of the buffer already prepared, calculate the pH when 1.8mL of 0.105M perchloric acid is added

Answer 1

a)

after mixing VT = V1+V2 = 160+140 = 300

pH = pKa + log(formate / formic acid)

pKa = -logKA = -log(1.8*10^-4) = 3.744

[formate ] = MV/VT = 140*0.25/300 = 0.1166 M

[formic acid[ = MV/VT = 160*0.15/300 = 0.08 M

then

pH = 3.744+ log(0.1166 /0.08 ) = 3.90760

B)

when addich HClO4

mmol of HClO4 = 1.8*0.105 = 0.189 mmol of H+

then

mmol of formic acid =  0.08 *100 = 8

mmol of formate = 0.1166 *100 = 11.67

after addition of H, expec formation of acid, reaction of foramte

mmol of formic acid = 8 + 0.189 = 8.189

mmol of formate =11.67 -0.189 = 11.481

then

pH = pKa + log(formate / formic acid) = 3.744 + log(11.481/8.189) = 3.8907