Question

an unknown weak acid with a concentration of 0.089 m has a ph of 1.80. what is the $k_{a}$ of the weak acid?

Explanation:

Step1: Calculate \([H^{+}]\)

Given \(pH = 1.80\), use the formula \(pH=-\log[H^{+}]\). Then \([H^{+}]=10^{-pH}\).

$$ [H^{+}]=10^{- 1.80}=0.0158\space M $$

Step2: Set up the equilibrium expression

Let the weak acid be \(HA\). The dissociation reaction is \(HA
ightleftharpoons H^{+}+A^{-}\). At equilibrium, \([H^{+}]=[A^{-}]=0.0158\space M\) and \([HA]=0.089 - 0.0158\space M\approx0.0732\space M\) (since \(0.089\gg0.0158\) for a weak acid).
The acid - dissociation constant \(K_{a}=\frac{[H^{+}][A^{-}]}{[HA]}\)

Step3: Calculate \(K_{a}\)

Substitute the values into the \(K_{a}\) formula:

$$ K_{a}=\frac{(0.0158)(0.0158)}{0.0732}=\frac{0.00024964}{0.0732}\approx3.41\times 10^{-3} $$

Answer:

\(K_{a}\approx3.41\times 10^{-3}\)