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What is the Ka of a 0.0796 M solution of nitrous acid (HNO2) with a pH of 2.95?

What is the Ka of a 0.0796 M solution of nitrous acid (HNO2) with a pH of 2.95?-example-1

1 Answer

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Answer:

Coefficient = 1.58

Exponent = - 5

Step-by-step explanation:

pH = 2.95

Molar concentration = 0.0796M

Ka = [H+]^2 / [HA]

Ka = [H+]^2 / 0.0796

Therefore ;

[H+] = 10^-2.95

[H+] = 0.0011220 = 1.122 × 10^-3

Ka = [H+] / molar concentration

Ka = [1.122 × 10^-3]^2 / 0.0796

Ka = (1.258884 × 10^-6) / 0.0796

Ka = 15.815 × 10^-6

Ka = 1.58 × 10^-5

Coefficient = 1.58

Exponent = - 5

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