Question

A 0.200 M solution of a weak acid, HA, is 9.4% ionized. Using this information, calculate Ka for HA.

Answer 1

${ \qquad\qquad\huge\underline{{\sf Answer}}}$

Let's solve ~

Initial concentration of weak acid HA = 0.200 M

and dissociation constant (${ \alpha}$) is :

$\qquad \sf \dashrightarrow \: \alpha = \frac{dissociation \: \: percentage}{100}$

$\qquad \sf \dashrightarrow \: \alpha = \frac{9.4}{100} = 0.094$

Now, at initial stage :

• $\textsf{ Conc of HA = 0.200 M}$

• $\textsf{Conc of H+ = 0 M}$

• $\textsf{Conc of A - = 0 M}$

At equilibrium :

• $\textsf{Conc of HA = 0.200 - 0.094(0.200) = 0.200(1 - 0.094) = 0.200(0.906) = 0.1812 M}$

• $\textsf{Conc of H+ = 0.094(0.200) = 0.0188 M}$

• $\textsf{Conc of A - = 0.094(0.200) = 0.0188 M}$

Now, we know :

$\qquad \sf \dashrightarrow \: { K_a = \dfrac{[H+] [A-]}{[HA]}}$

( big brackets represents concentration )

$\qquad \sf \dashrightarrow \: { K_a = \dfrac{0.0188×0.0188}{0.1812}}$

$\qquad \sf \dashrightarrow \: { K_a = \dfrac{0.00035344}{0.1812}}$

$\qquad \sf \dashrightarrow \: { K_a \approx 0.00195 }$

$\qquad \sf \dashrightarrow \: {K_a \approx 1.9 × {10}^{-3} }$