Question

In: Chemistry

You wish to titrate a 0.164 M solution of HCN (pKa = 9.30) with a 0.161...

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?

Solutions

Expert 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


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