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Enough of a monoprotic acid is dissolved in water to produce a 1. 50 m solution. the ph of the resulting solution is 2. 83. calculate the ka for the acid

Sagot :

1.4406 × 10⁻⁶ is the Ka(dissociation constant) for the acid.

The equilibrium constant for the reaction of an acid with water is the acid dissociation constant, where the acid, HA separates into H⁺ and A⁻ ions.

The acid dissociation constant is represented by (Ka).

So let's first imagine that the given monoprotic acid is HA.

HA will dissociate into H⁺ and A⁻ ions.

HA ⇒ H⁺ and A⁻

The formula used for Ka is

Ka = [H⁺] [A⁻] / [HA]

Given

pH = 2.83

[HA] = 1.50 M

From the given pH, we can calculate [H⁺] and [A⁻]

[H⁺] = [A⁻] = 1 × [tex]10^{-2.83}[/tex] =  1.47 × 10⁻³ M

Ka = 1.47 × 10⁻³ × 1.47 × 10⁻³ / 1.50

Ka = 1.4406 × 10⁻⁶

Hence, 1.4406 × 10⁻⁶ is the Ka for the acid.

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