To what volume (in litres) must one litre of a $1.0 ext{ M}$ solution of acetic acid be diluted so

Question

To what volume (in litres) must one litre of a $1.0	ext{ M}$ solution of acetic acid be diluted so that the pH of the resulting solution will be twice the original value? Given: $K_a = 1.8 	imes 10^{-5}$.

Professor PC Thomas Classes · Answer

Let the original pH be $	ext{pH}_1$. As calculated, $[	ext{H}^+]_1 = \sqrt{K_a C_1}$.
We want the new pH, $	ext{pH}_2$, to be $2 	imes 	ext{pH}_1$.
This means $-\log[	ext{H}^+]_2 = -2\log[	ext{H}^+]_1 \implies [	ext{H}^+]_2 = ([	ext{H}^+]_1)^2$.
We know $[	ext{H}^+]_1 = \sqrt{K_a C_1}$. So $[	ext{H}^+]_2 = K_a C_1$.
Since $C_1 = 1.0	ext{ M}$, $[	ext{H}^+]_2 = K_a$.
For the diluted solution, let the new concentration be $C_2$. The new hydrogen ion concentration is given by the exact formula if dissociation is significant, or the approximate one. However, if $[	ext{H}^+]_2 = K_a$, we can use the equilibrium expression:
$K_a = \frac{[	ext{H}^+][	ext{Ac}^-]}{[	ext{HAc}]} \approx \frac{[	ext{H}^+]^2}{C_2 - [	ext{H}^+]}$
Substitute $[	ext{H}^+]_2 = K_a$:
$K_a = \frac{K_a^2}{C_2 - K_a} \implies 1 = \frac{K_a}{C_2 - K_a} \implies C_2 - K_a = K_a \implies C_2 = 2K_a$.
Given $K_a = 1.8 	imes 10^{-5}$, $C_2 = 3.6 	imes 10^{-5}	ext{ M}$.
The dilution formula is $M_1 V_1 = M_2 V_2$.
$1.0 	imes 1 = (3.6 	imes 10^{-5}) 	imes V_2$.
$V_2 = \frac{1}{3.6 	imes 10^{-5}} = \frac{10^5}{3.6} = 27777.78	ext{ L}$.
Thus, the volume is approximately $2.78 	imes 10^4	ext{ L}$. Rounding to nearest integer gives $27778$.