Question

What would be the ?nal pH if 0.0100 moles of solid NaOH were added to 100 mL of a bu?er solution containing 0.600 molar formic acid (ionization constant = 1.8 × 10?4) and 0.300 molar sodium formate

Answer 1

Moles formic acid initially present in a given 100 mL buffer = 100 mL x 10^-3 L x 0.6 M = 0.06 mol

Moles sodium formate initially present in a given 100 mL buffer = 100 mL x 10^-3 L x 0.3 M = 0.03 mol

Since we are adding a base hence it will reacts with acid only and forms more sodium formate.

R..................HCOOH (aq) + OH- (aq) <==> HCOO- (aq)

I................... 0.06................0.01.................0.03

C..................-0.01...............-0.01.................+0.01

E...................0.05................0.00..................0.04

given Ka = 1.8 x 10^-4 ==> pKa = -logKa = -log[1.8 x 10^-4] = 3.74

from Henderson-Hasselbalch equation for acidic buffers we have

pH = pKa + log[salt/acid]

pH = 3.74 + log[0.04 / 0.05] = 3.74 - 0.09 = 3.65

pH = 3.65 ----------------------- correct answer.

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