Question

Enough of a monoprotic weak acid is dissolved in water to produce a 0.01660.0166 M solution. The pH of the resulting solution is 2.532.53 . Calculate the Ka for the acid.

Answer 1

Step-by-step explanation:

Let the monoprotic acid be HX

HX ⇄ H⁺ + X⁻

pH = 2.53

Hydrogen ion concentration

`[ H^+]=10^(-2.53)`

`[ X^-]=10^(-2.53)`

Concentration of undissociated acid will remain almost the same as it is a weak acid

So

Ka = concentration of H⁺ x concentration of Cl⁻ / concentration of acid

= [ H⁺] x [Cl⁻ ] / [ HX]

`k_a=(10^(-2.53)* 10^(-2.53))/(.0166)`

`k_a=(.00295^2)/(.0166)`

= 5.24 x 10⁻⁴ M .