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

Answer:

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]

$k_a=\frac{10^{-2.53}\times 10^{-2.53}}{.0166}$

$k_a=\frac{.00295^2}{.0166}$

= 5.24 x 10⁻⁴ M .