Buffer Question : addition of base

Question

Calculate the pH of the solution in part A after 25mL of 0.10 M HBr are added to 125 mL of the buffer solution (Part A : 0.35 M HCN, 0.37 M NaCN, ka = 6.2e-10, pH calculated as 9.23)

Answer 1

From part A, the pH is calculated to be 9.23. As a check, I will find it here using HH equation where pH = pKa + log [salt]/[acid]:

pH = 9.21 + 0.02 = 9.23 (check)

125 ml buffer x 1L/1000 ml x 0.37 mol/L NaCN = 0.04625 moles NaCN

125 ml buffer x 1L/1000 ml x 0.35 mol/L HCN = 0.04375 moles HCN

25 ml HBr x 1L/1000 ml x 0.10 mol/L HBr = 0.0025 moles HBr added

When HBr is added it reacts with the NaCN to reduce the moles of NaCN and increase the moles of HCN

HBr + NaCN ==> HCN + NaBr

Moles of HCN formed = 0.0025

Final moles HCN = 0.04375 + 0.0025 = 0.04625 moles HCN

Final moles NaCN = 0.04625 - 0.0025 = 0.04375 moles NaCN

Final volume = 125 ml + 25 ml = 150 ml = 0.150 L

Final [HCN] = 0.04625 moles/0.150 L = 0.3083 M

Final [NaCN] = 0.04375 moles/0.150 L = 0.2917 M

Final pH = pKa + log [NaCN]/[HCN] = 9.21 + log (0.2917/0.3083) = 9.21 + (-0.02) = 9.19