In: Chemistry
You wish to titrate a 0.164 M solution of HCN (pKa = 9.30) with a 0.161 M solution of NaOH. If you start with 44.8 mL of the HCN solution, what is the pH after the addition of 10.1 mL NaOH solution?
Before addition of NaOH
Concentration of HCN =C = 0.164 M
Moles of HCN = 0.164 x 0.0448 = 0.00735 mol
Finding equilibrium concentrations of ions before addition of NaOH
HCN + H2O <===> H3O+ + CN-
Concentration of H3O+ ions =
Given pKa = 9.30
Ka = 10-pKa
Ka = 5.01 x 10-10
[H3O+] = [CN-]= 9.07 x 10-6 M
moles of H3O+ ions = Molarity x volume = 9.07 x 10-6 x 0.0448 = 4.06 x 10-7 mol = moles of CN- ions
After addtion of 10.1 mL of 0.161 M NaOH
Moles of NaOH = 0.161 x 0.0101 = 0.00162 mol
HCN | OH- | CN- | H2O | |
Initial, moles | 0.00735 | 0.00162 | 4.06 x 10-7 | 4.06 x 10-7 |
change | -0.00162 | -0.00162 | +0.00162 | +0.00162 |
final | 0.00573 | 0 | 0.00162 | 0.00162 |
Using Henderson-Hasselbalch equation
Therefore, pH = 8.75