Question #27014
Calculate the pH of the following solution:
0.1 M HF + 0.2 M NaCN
Given: pKa for: HF/F- = 3.2 HCN/CN-= 9.4
0.1 M HF + 0.2 M NaCN
Given: pKa for: HF/F- = 3.2 HCN/CN-= 9.4
Expert's answer
The chemical equation for this reaction is
HF + NaCN = NaF + HCN
HCN is much more weak acid than HF because it's pKa is greater
If the volumes of HF and NaCN were equal, HF is limiting reagent and is completely transformed into NaF. In the solution there is buffer HCN/CN-, and the pH of this buffer is
pH = pKa + lg[A-]/[HA]
pH = 9.4 + lg([CN-]/([HCN]) = 9.4 + lg (0.1/0.1) = 9.4
Answer: pH = 9.4
HF + NaCN = NaF + HCN
HCN is much more weak acid than HF because it's pKa is greater
If the volumes of HF and NaCN were equal, HF is limiting reagent and is completely transformed into NaF. In the solution there is buffer HCN/CN-, and the pH of this buffer is
pH = pKa + lg[A-]/[HA]
pH = 9.4 + lg([CN-]/([HCN]) = 9.4 + lg (0.1/0.1) = 9.4
Answer: pH = 9.4
