Question

On the basis of the information above, what is the approximate percent ionization of HNO2 in a 1.0 M HNO2 (aq) solution?

Answer 1

Answer:

The answer is "2%"

Explanation:

Equation:

$HNO_2\ (aq) \leftrightharpoons H^{+} \ (aq) + NO_2^{-}\ (aq) \\\\\ K_a = 4.0\times \ 10^{-4}$

$H^{+}=?$

Formula:

$Ka = \frac{[H^{+}][NO_2^{-}]}{[HNO_2]}$

Let

$[H^{+}] = [NO_2^{-}] = x$ at equilibrium

$x^2 = (4.0\times 10^{-4})\times 1.0\\\\x = ((4.0\times 10^{-4})\times 1.0)^{0.5} = 2.0 \times 10^{-2} \ M$

therefore,

$[H^{+}] = 2.0\times 10^{-2} \ M = 0.02 \ M$

Calculating the % ionization:

$= \frac{([H^{+}]}{[HNO_2])} \times 100 \\\\= \frac{0.02}{1}\times 100 \\\\= 2\%$