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:

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

Answer 2

The approximate percent ionization of HNO₂ in a 1.0 M HNO₂ (aq) solution is 2%.

How we calculate the % ionization?

% ionization of any compound will be calculated as follow:

% ionization = ([ion]/[acid or base]) ₓ 100

Given chemical reaction with ICE table will be represented as:

                           HNO₂(aq) → H⁺(aq) + NO₂⁻(aq)

initial                          1                0             0

change                      -x               +x           +x

equilibrium              1-x                x             x

Equilibrium constant will be represented as:
Ka = [H⁺][NO₂⁻] / [HNO₂]

Acid dissociation constant for HNO₂ = 4×10⁻⁴

Putting all values in the above equation, we get

4×10⁻⁴ = x² / 1-x

Value of changeable quantity is very less, so we neglect from the concentration of HNO₂.

4×10⁻⁴ = x²

x = 2 × 10⁻²

So, the concentration of H⁺ ion = 2 × 10⁻²M

Now we put all these values in the % ionization equation, we get

% ionization = (0.02/1) × 100 = 2%

Hence , % ionization is 2%.

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