Question

the $k_{eq}$ for the reaction of water with water is called $k_w$
$h_2o + h_2o \leftrightarrow h_3o^{+1} + oh^{-1}$
\qquad\quad $\frac{h_3o^{+1}oh^{-1}}{h_2o^2}$
$k_w = \frac{h_3o^{+1}oh^{-1}}{h_2o^2} = h_3o^{+1}oh^{-1} = (1 \times 10^{-7})^2 = 1 \times 10^{-14}$
the inverse logarithm, or -log, of $h_3o^{+1}oh^{-1} = 10^{-14}$ is $ph + poh = 14$

1. what happens to the $h_3o^{+1}$ when acid is added to neutral water ?

\quad the $h_3o^{+1}$ goes 1

1. what happens to the $ph$ when acid is added to neutral water ?

\quad an example of an increased $h_3o^{+1}$ is $h_3o^{+1} = 1 \times 10^{-6}$ and $ph = -\log10^{-6} = 6$
\quad thus, adding acid makes the $ph$ go 2

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

\quad in the equation $h_3o^{+1}oh^{-1} = 10^{-14}$, when $h_3o^{+1}$ increases, the $oh^{-1}$ goes 3

1. what happens to the $poh$ when acid is added to neutral water ?

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

Explanation:

Question 1

Step1: Recall acid properties

Acids donate \(H^+\) ions, which combine with \(H_2O\) to form \(H_3O^+\).

Step2: Analyze concentration change

Adding acid increases the source of \(H^+\), so \( [H_3O^{+1}] \) increases.

Step1: Recall pH formula

\( pH = -\log[H_3O^{+1}] \).

Step2: Relate to concentration change

When \( [H_3O^{+1}] \) increases (from acid addition), the log of a larger number is larger, but with a negative sign, so \( pH \) decreases (e.g., \( 10^{-7} \) to \( 10^{-6} \) makes pH go from 7 to 6).

Step1: Use \( K_w \) relationship

\( K_w = [H_3O^{+1}][OH^{-1}] = 1\times10^{-14} \) (constant at 25°C).

Step2: Analyze inverse relationship

If \( [H_3O^{+1}] \) increases, to keep \( K_w \) constant, \( [OH^{-1}] \) must decrease.

Answer:

up (or increases)

Question 2