Question

Enough of a monoprotic acid is dissolved in water to produce a 0.0116 M solution. The pH of the resulting solution is 2.35. Calculate the Ka for the acid.

Answer 1

`K_(a)` = 1.72 × 10⁻³

HA + H₂O ⇌ H₃O⁺ + A⁻


`K_(a) = ([H_(3) O^(+)][ A^(-)] )/([HA])`

[HA] = 0.0116 mol/L

pH = 2.35

[H₃O⁺] =
`10^(-pH)` mol/L=
`10^(-2.35)` mol/L = 4.47 × 10⁻³ mol/L

[A⁻] = [H₃O⁺] = 4.47 × 10⁻³ mol/L


`K_(a) = ([H_(3) O^(+)][ A^(-)] )/([HA])` =
`(4.47 * 10^(-3) * 4.47 * 10^(-3))/( 0.0116)` = 1.72 × 10⁻³