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: K_a (HNO₂/NO₂^-) = 5.62 x 10^-4

Question

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: K_a (HNO₂/NO₂^-) = 5.62 x 10^-4

| General Chemistry 3

Accepted Answer

Henderson-Hasselbalch equation can be used:

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

pH = 3.25 + log

\frac{{[{NO}_{2}^{-}]}_{0}}{{[{HNO}_{2}]}_{0}}

\frac{{[{NO}_{2}^{-}]}_{0}}{{[{HNO}_{2}]}_{0}}

= 10^0.45

\frac{{[{NO}_{2}^{-}]}_{0}}{{[{HNO}_{2}]}_{0}}

=

\frac{n_{{NO}_{2}^{-}}}{V \times {[{HNO}_{2}]}_{0}}

=

\frac{m_{{NaNO}_{2}}}{M_{{NaNO}_{2}} \times V \times {[{HNO}_{2}]}_{0}}

⇒ m_NaNO2 = 10^0.45 x M_NaNO2 V x [HNO₂]₀

⇒ m_NaNO2 = 97.2 g

Exercise 9 | Buffers and Acid-Base Titrations