Chapter 16 - Acids and Bases - Exercises - Weak Acids, Weak Bases, and pH - Page 740: 37

(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.

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