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#)?

Question

Serenity Allen · Accepted Answer

Well the buffer equation operates....

It states that... #pH=pK_a+log_10{[[A^-]]/[[HA]]}# and so here....#pH=3.15+log_10((0.68*mol*L^-1)/(0.24*mol*L^-1))=3.15+0.453=3.60# And thus... #[H_3O^+]=10^(-3.60)*mol*L^-1=2.51xx10^-4*mol*L^-1#. Solution #pH# is raised relative to #pK_a# given that the concentration of the conjugate base, the nitrite anion, is GREATER than that of the nitrous acid. What would #pH# be, should #[NO_2^(-)]=[HNO_2]#?

Avery Adams · Answer

undefined To find the $ \text{H}_3\text{O}^+ $ concentration and the pH of the buffer, we first need to determine the moles of $ \text{HNO}_2 $ and $ \text{NO}_2^- $ in the buffer solution.

Given:

- $ \text{[HNO}_2\text{]} = 0.24 $ M
- $ \text{[KNO}_2\text{]} = 0.68 $ M

The buffer reaction is:

$$ \text{HNO}_2 \rightleftharpoons \text{H}^+ + \text{NO}_2^- $$

Using the Henderson-Hasselbalch equation:

$$ \text{pH} = \text{p}K_a + \log\left( \frac{\text{[A^-]}}{\text{[HA]}} \right) $$

where $ \text{[A^-]} $ is the concentration of the conjugate base ($ \text{NO}_2^- $) and $ \text{[HA]} $ is the concentration of the weak acid ($ \text{HNO}_2 $), and $ \text{p}K_a = -\log K_a $.

First, calculate the ratio $ \frac{\text{[A^-]}}{\text{[HA]}} $:

$$ \frac{0.68}{0.24} = 2.83 $$

Then, find the pH:

$$ \text{pH} = -\log(7.1 \times 10^{-4}) + \log(2.83) $$

$$ \text{pH} = 3.15 + 0.45 $$

$$ \text{pH} = 3.60 $$

Thus, the pH of the buffer is 3.60.