Question

If 0.15M solution of HNO2 has a pH of 1.98, determine the % ionization of HNO2.

Answer 1

The % ionization of HN0₂ is 6.98 %

What is percentage Ionization ?

The relative strengths of acids may be determined by measuring their equilibrium constants in aqueous solutions.

In solutions of the same concentration, stronger acids ionize to a greater extent, and so yield higher concentrations of hydronium ions than do weaker acids.

Percentage Ionization is the measure of the strength of an acid is its percent ionization.

The percent ionization of a weak acid is the ratio of the concentration of the ionized acid to the initial acid concentration, times 100:

The chemical equation for the dissociation of the nitrous acid is:

HNO₂(aq)+H₂O(l)⇌NO²⁻(aq)+H₃O⁺(aq).

$\rm \% \;ionization = \dfrac{[H_{3}O^{+} ] _{eq} }{[HNO_{2}]_{0} } \times 100$

 $\rm 10^{-pH} =$ [H₃O⁺] ,

we find that with given pH of 1.98

$10^{-1.98} =$ 0.01047128548 M=[H₃O⁺]

Molarity of HN0₂ = 0.15 M

so that percent ionization is

$\% ionization = \dfrac{0.01047128548 }{0.15 } \times 100$

$\% ionization = 6.98 \%$

Therefore the % ionization of HN0₂ is 6.98 %

To know more about percentage Ionization

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