## General Chemistry: Principles and Modern Applications (10th Edition)

Published by Pearson Prentice Hal

# Chapter 16 - Acids and Bases - Exercises - Weak Acids, Weak Bases, and pH - Page 740: 38

#### Answer

$(c)$ is the one that better represents this solution.

#### Work Step by Step

1. Considering a pure solution: $[OH^-] = [B^+] = x$ 2. Use the Kb formula: $Kb = \frac{[OH^-][B^+]}{[Base]} = \frac{x^2}{[Base]}$ 3. If the solution is diluted to half, the $OH^-$ will change, the base will half its concentration, but the Kb will still stay the same: $Kb = \frac{y^2}{[Base]*0.5}$ 4. Using algebra to solve: $Kb = \frac{x^2}{[Base]} = \frac{y^2}{[Base]0.5}$ - We can eliminate the [Base] $x^2 = \frac{y^2}{0.5}$ $0.5 * x^2 = y^2$ - We can use square roots to solve: $\sqrt {0.5 * x^2} = \sqrt {y^2}$ $0.7 * x \approx y$ - So the second solution will have 0.7 times the concentration of the first one. - The image of the first solution has 25 protons, so the second should have 25 * 0.7 = 17.5. (c) has 17 protons, so it is the closest one.

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