Answer
(b) is the sketch that better represents.
Work Step by Step
1. Considering a pure solution:
$[H_3O^+] = [A^-] = x$
Therefore:
$Ka = \frac{[H_3O^+][A^-]}{[HA]} = \frac{x^2}{[HA]}$
2. If the molarity of the solution is doubled:
$Ka = \frac{y^2}{2[HA]}$
** The ka is constant, but the $[H_3O^+]$ changes, so we consider other unknown : y.
Since the ka is constant, we can say that:
$\frac{x^2}{[HA]} = \frac{y^2}{2[HA]}$
- We can eliminate the $[HA]$:
$x^2 = \frac{y^2}{2}$
- And put a square root in both sides:
$\sqrt {x^2} = \sqrt { \frac{y^2}{2}}$
$x \approx \frac{y}{1.4}$
$y \approx 1.4x$
- Since x is the concentration of the far left $H_3O^+$, and it has 10 protons in the image:
$y = 10 * 1.4$
$y= 14$
Then "y" should have 14 protons in its image, so (b) is the right answer.
