Question

A buffer solution contains 0.384 M KHCO3 and 0.239 M Na2CO3. If 0.0464 moles of potassium hydroxide are added to 225.0 ml of this buffer, what is the PH of the resulting soulution?

(Assune the volume does not change)

I saw a few others posted this question on here, but is there a way to do this problem without using the Ka of HCO3? Either way itd be great to see how to do this problem. Thanks!

Answer 1

pH of buffer = pKa + log ([salt]/[acid]) = pKa + log (0.239/0.384) = pKa - 0.206 = 10.32 - 0.206 = 10.114

KOH + KHCO₃ $\rightarrow$ K₂CO₃ + H2O

mole of KHCO₃ = 225.0×10^-3 × 0.384= 0.0864

Moles of KOH added = 0.0464

Moles of KHCO3 left = 0.0864 - 0.0464 = 0.0400

[KHCO3] = 0.0400/(0.225) = 0.178M

Moles of CO₃^2- = 0.239 × 0.225 + 0.0464 = 0.100

pH of new buffer = pKa + log (0.100/0.0400) = pKa + 0.399

pKa of HCO₃^- = 10.32 (Ka= 4.8×10^-11)

pH = 10.32 + 0.399 = 10.719 = 10.72

For calculating pH of solution we require Ka of acid HCO₃^- . We can calculate change in pH after adding KOH, because in that case pKa in two equations will cancel out each other.

Change in pH (∆pH) = pka + 0.399 - (pKa - 0.206) = 0.605