# Exercise 9 | Buffers and Titration of Acids and Bases

Calculate the mass of NaNO2 (s) that must be added to 1.00 L of 0.500 M HNO2 (aq) to give a pH of 3.70

Data: Ka (HNO2/NO2-) = 5.62 x 10-4

Henderson-Hasselbalch equation can be used:

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

pH = 3.25 + log $\frac{{\left[{{\mathrm{NO}}_{2}}^{-}\right]}_{0}}{{\left[{\mathrm{HNO}}_{2}\right]}_{0}}$

$\frac{{\left[{{\mathrm{NO}}_{2}}^{-}\right]}_{0}}{{\left[{\mathrm{HNO}}_{2}\right]}_{0}}$ = 100.45

$\frac{{\left[{{\mathrm{NO}}_{2}}^{-}\right]}_{0}}{{\left[{\mathrm{HNO}}_{2}\right]}_{0}}$=  =

⇒ mNaNO2 = 100.45 x MNaNO2 V x [HNO2]0

⇒ mNaNO2 = 97.2 g