# What is the H_3O^+ and the pH of a buffer that consists of 0.24 M HNO_2 and 0.68 M KNO_2? (K_a of HNO_2 = 7.1 x 10^-4)?

Jan 3, 2018

#### Explanation:

...which says that...

$p H = p {K}_{a} + {\log}_{10} \left\{\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right\}$

and so here....$p H = 3.15 + {\log}_{10} \left(\frac{0.68 \cdot m o l \cdot {L}^{-} 1}{0.24 \cdot m o l \cdot {L}^{-} 1}\right) = 3.15 + 0.453 = 3.60$

And so.....

$\left[{H}_{3} {O}^{+}\right] = {10}^{- 3.60} \cdot m o l \cdot {L}^{-} 1 = 2.51 \times {10}^{-} 4 \cdot m o l \cdot {L}^{-} 1$.

Solution $p H$ is raised relative to $p {K}_{a}$ given that the concentration of the conjugate base, the nitrite anion, is GREATER than that of the nitrous acid.

What would $p H$ be, should $\left[N {O}_{2}^{-}\right] = \left[H N {O}_{2}\right]$?