Question

1. a saturated solution of caproic acid, hc₆h₁₁o₂, contains 11 g of the acid/l solution. the ph of the solution is 2.94. what is the kₐ and pkₐ of the acid?

Explanation:

Step1: Calculate the initial concentration of caproic acid

The molar mass of $HC_6H_{11}O_2$ is $M=(1\times1 + 6\times12+11\times1 + 2\times16)=116\ g/mol$. Given the mass - per - volume concentration $m = 11\ g/L$, the initial concentration $c$ of the acid is $c=\frac{11\ g/L}{116\ g/mol}=0.0948\ mol/L$.

Step2: Determine the concentration of $H^+$ ions

We know that $pH = -\log[H^+]$. Given $pH = 2.94$, then $[H^+]=10^{-pH}=10^{- 2.94}=1.15\times10^{-3}\ mol/L$.
For a weak acid $HA
ightleftharpoons H^++A^-$, at equilibrium, $[H^+]=[A^-]$ and $[HA]=c - [H^+]$. Since $c = 0.0948\ mol/L$ and $[H^+]=1.15\times10^{-3}\ mol/L$, and $c\gg[H^+]$, we can approximate $[HA]\approx c$.

Step3: Calculate the acid - dissociation constant $K_a$

The expression for the acid - dissociation constant of a weak acid $HA$ is $K_a=\frac{[H^+][A^-]}{[HA]}$. Since $[H^+]=[A^-]=1.15\times10^{-3}\ mol/L$ and $[HA]\approx0.0948\ mol/L$, then $K_a=\frac{(1.15\times10^{-3})\times(1.15\times10^{-3})}{0.0948}=1.4\times10^{-5}$.

Step4: Calculate the $pK_a$

The relationship between $pK_a$ and $K_a$ is $pK_a=-\log K_a$. So $pK_a=-\log(1.4\times10^{-5}) = 4.85$.

Answer:

$K_a = 1.4\times10^{-5}$, $pK_a=4.85$