# Exercise 3 | Buffers and Titration of Acids and Bases

Calculate the pH of a solution originally containing 0.200 mol of cyanic acid HCNO following addition of 80.0 mL of 1.00 M NaOH.

Data: Ka (HCNO/-CNO) = 3.5 x 10-4

Reaction: HCNO + HO--CNO + H2O

Calculate the number of moles of -CNO formed:

nCNO = nHO = 8.00 x 10-2 x 1.00 = 8.00 x 10-2 mol

nHCNO = nHCNO-initial - nHO = 1.60 x 10-1 mol

Henderson-Hasselbalch equation can be used:

pH = pKa + log $\frac{{\left[\mathrm{base}\right]}_{0}}{{\left[\mathrm{acid}\right]}_{0}}$

pH = pKa + log $\frac{{\mathrm{n}}_{\mathrm{base}}}{{\mathrm{n}}_{\mathrm{acid}}}$

pH = 3.28