5. the ph increases by 2 units. how does h⁺ change?\n6. the ph decreases by 1 unit. how does h⁺ change?

Explanation:

Step1: Recall the pH formula

The formula for pH is $pH = -\log[H^{+}]$. Let the initial pH be $pH_1=-\log[H^{+}_1]$ and the final pH be $pH_2 = -\log[H^{+}_2]$.

Step2: Solve for problem 5

Given $pH_2 - pH_1=2$. Substitute the pH - formula: $(-\log[H^{+}_2])-(-\log[H^{+}_1]) = 2$. Using the logarithmic property $\log a-\log b=\log\frac{a}{b}$, we get $\log\frac{[H^{+}_1]}{[H^{+}_2]} = 2$. By the definition of logarithms, if $\log x = y$, then $x = 10^{y}$. So, $\frac{[H^{+}_1]}{[H^{+}_2]}=10^{2}=100$, and $[H^{+}_2]=\frac{[H^{+}_1]}{100}$. The hydrogen - ion concentration decreases by a factor of 100.

Step3: Solve for problem 6

Given $pH_1 - pH_2 = 1$. Substitute the pH - formula: $(-\log[H^{+}_1])-(-\log[H^{+}_2])=1$. Using the logarithmic property $\log a-\log b=\log\frac{a}{b}$, we get $\log\frac{[H^{+}_2]}{[H^{+}_1]} = 1$. By the definition of logarithms, if $\log x = y$, then $x = 10^{y}$. So, $\frac{[H^{+}_2]}{[H^{+}_1]}=10^{1}=10$, and $[H^{+}_2]=10[H^{+}_1]$. The hydrogen - ion concentration increases by a factor of 10.

Answer:

5. The $[H^{+}]$ decreases by a factor of 100.
6. The $[H^{+}]$ increases by a factor of 10.