Question

If the a of a monoprotic weak acid is 1. 2×10^−6. What is the [H] of a 0. 31 m solution of this acid?

Answer 1

The pH of the solution of the given acid is 3.099

Let  HX  be the weak acid:

HX ⇌  $H^{+}$+ $X^{-}$

For which:

$K_{a}$ = [ $H^{+}$][ $X^{-}$] / [HX]

If  $K_{a}$ lies in the range $10^{-4} - 10^{-10}$we can assume that the equilibrium concentrations used in the expression are a good enough approximation to the initial concentrations.

Rearranging and taking negative logs of both sides gives:

pH = $\frac{1}{2} [pK_{a} - loga]$

a is the concentration of the acid.

$pK_{a} =-logK_{a}$ = -log (2×$10^{-6}$) = 5.69

pH = $\frac{1}{2} [5.69- (-0.508)]$

pH = 3.099

Learn more about pH of acid here;

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