Question

he weak acid HQ has a pK_a of 4.89. Calculate the [H₃O⁺ ] of 0.020 M HQ.

Answer 1

$pKa =-\log(Ka)\\ \log(Ka) = -pKa\\ Ka=10^{-pKa}\\ Ka= 10^{-4.89}\\ Ka = 1.288×10^{-5}$

\begin{alignedat}{4} &HQ_{(aq)}\quad + &\quad H_2O_{(l)}\quad _\longrightarrow^\longleftarrow\quad &H_3O^+_{(aq)}\quad + &\quad Q^-_{(aq)} \end{alignedat}

$\begin{aligned} &\textsf{Initial Conc. : }0.020M +\ \text{---}M\ &= 0M + 0M\\ &\textsf{Eqm: }(0.020-x)M +\ \text{---}M\ &= xM + xM \end{aligned}$

$\begin{aligned} Ka = \dfrac{products}{reactants} = \dfrac{[H_3O^+][Q^-]}{HQ} \end{aligned}$

$\begin{aligned} 1.288 × 10^{-5} = \dfrac{(x)(x)}{(0.020-x)} \end{aligned}$

Since x < < 1,

0.020 - x $\approx$ 0.020

$\begin{aligned} 1.288 × 10^{-5} = \dfrac{x^2}{(0.020)} \end{aligned}$

$\begin{aligned} x^2 = 1.288 × 10^{-5} × 0.020 \end{aligned}$

$\begin{aligned} x = \sqrt{1.288 × 10^{-5} × 0.020\ } \end{aligned}$

$x = 1.605 × 10^{-3} M$

$\begin{aligned} [H_3O^+] = x = 1.605 × 10^{-3}M \end{aligned}$