Exercise 3 | Buffers and Titration of Acids and Bases

Reaction: HCNO + HO^- → ^-CNO + H₂O

Calculate the number of moles of ^-CNO formed:

n_CNO = n_HO = 8.00 x 10^-2 x 1.00 = 8.00 x 10^-2 mol

n_HCNO = n_HCNO-initial - n_HO = 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