Question

the $k_{eq}$ for the reaction of water with water is called $k_w$
$\ce{h2o + h2o <=> h3o^{+1} + oh^{-1}}$
\\k_w = \frac{\ce{h3o^{+1}} \ce{oh^{-1}}}{\ce{h2o}^2} = \ce{h3o^{+1}} \ce{oh^{-1}} = (1 \times 10^{-7})^2 = 1 \times 10^{-14}\\
the inverse logarithm, or -log, of $\ce{h3o^{+1}} \ce{oh^{-1}} = 10^{-14}$ is $\ce{ph + poh = 14}$

1. what happens to the $\ce{h3o^{+1}}$ when acid is added to neutral water?

the $\ce{h3o^{+1}}$ goes 1

1. what happens to the $\ce{ph}$ when acid is added to neutral water?

an example of an increased $\ce{h3o^{+1}}$ is $\ce{h3o^{+1}} = 1 \times 10^{-6}$ and $\ce{ph} = -\log_{10}10^{-6} = 6$
thus, adding acid makes the $\ce{ph}$ go 2

1. what happens to the $\ce{oh^{-1}}$ when acid is added to neutral water?

in the equation $\ce{h3o^{+1}} \ce{oh^{-1}} = 10^{-14}$, when $\ce{h3o^{+1}}$ increases, the $\ce{oh^{-1}}$ goes 3

1. what happens to the $\ce{poh}$ when acid is added to neutral water?

when $\ce{oh^{-1}}$ decreases, the $\ce{poh}$ goes 4 since the $\ce{poh} = -\log\ce{oh^{-1}}$
this can also be seen using $\ce{ph + poh = 14}$

Explanation:

Step1: Analyze [H₃O⁺] change

Acids donate H⁺ ions. In water, H⁺ combines with H₂O to form H₃O⁺. So adding acid increases [H₃O⁺]. So 1 is "up" (or "increase").

Step2: Analyze pH change

pH = -log[H₃O⁺]. As [H₃O⁺] increases (e.g., from \(1\times10^{-7}\) to \(1\times10^{-6}\)), -log of a larger number (closer to 0) decreases. So pH goes down. 2 is "down" (or "decrease").

Step3: Analyze [OH⁻] change

Given \(K_w = [H_3O^+][OH^-] = 1\times10^{-14}\) (constant at 25°C). If [H₃O⁺] increases, to keep \(K_w\) constant, [OH⁻] must decrease. So 3 is "down" (or "decrease").

Step4: Analyze pOH change

pOH = -log[OH⁻]. If [OH⁻] decreases, -log of a smaller number (more negative) increases. For example, if [OH⁻] goes from \(1\times10^{-7}\) to \(1\times10^{-8}\), pOH goes from 7 to 8. So 4 is "up" (or "increase").

Answer:

1. up (increase)
2. down (decrease)
3. down (decrease)
4. up (increase)