h₂o + h₂o ↔ h₃o⁺¹ + oh⁻¹
h₃o⁺¹ oh⁻¹
k_w = ------------- = h₃o⁺¹ oh⁻¹ = (1 × 10⁻⁷)² = 1 × 10⁻¹⁴
h₂o²
the inverse logarithm, or -log, of h₃o⁺¹ oh⁻¹ = 10⁻¹⁴ is ph + poh = 14
1. what happens to the h₃o⁺¹ when acid is added to neutral water ?
the h₃o⁺¹ goes __1__
2. what happens to the ph when acid is added to neutral water ?
an example of an increased h₃o⁺¹ is h₃o⁺¹ = 1 × 10⁻⁶ and ph = -log10⁻⁶ = 6
thus, adding acid makes the ph go __2__
3. what happens to the oh⁻¹ when acid is added to neutral water ?
in the equation h₃o⁺¹ oh⁻¹ = 10⁻¹⁴, when h₃o⁺¹ increases, the oh⁻¹ goes __3__
4. what happens to the poh when acid is added to neutral water ?
when oh⁻¹ decreases, the poh goes __4__ since the poh = -logoh⁻¹
this can also be seen using ph + poh = 14
5. what happens to the k_w when acid is added to neutral water ?

Question

h₂o + h₂o ↔ h₃o⁺¹ + oh⁻¹
        h₃o⁺¹ oh⁻¹
k_w = ------------- = h₃o⁺¹ oh⁻¹ = (1 × 10⁻⁷)² = 1 × 10⁻¹⁴
        h₂o²
the inverse logarithm, or -log, of h₃o⁺¹ oh⁻¹ = 10⁻¹⁴ is ph + poh = 14
1. what happens to the h₃o⁺¹ when acid is added to neutral water ?
   the h₃o⁺¹ goes __1__
2. what happens to the ph when acid is added to neutral water ?
   an example of an increased h₃o⁺¹ is h₃o⁺¹ = 1 × 10⁻⁶ and ph = -log10⁻⁶ = 6
   thus, adding acid makes the ph go __2__
3. what happens to the oh⁻¹ when acid is added to neutral water ?
   in the equation h₃o⁺¹ oh⁻¹ = 10⁻¹⁴, when h₃o⁺¹ increases, the oh⁻¹ goes __3__
4. what happens to the poh when acid is added to neutral water ?
   when oh⁻¹ decreases, the poh goes __4__ since the poh = -logoh⁻¹
   this can also be seen using ph + poh = 14
5. what happens to the k_w when acid is added to neutral water ?

Accepted Answer

### Sub - question 1
# Explanation:
## Step1: Recall acid - water reaction
Acids dissociate in water to give $H_3O^+$ ions. For example, $HCl + H_2O
ightarrow H_3O^++Cl^-$.
## Step2: Analyze $ [H_3O^{+1}] $ change
When acid is added to neutral water, the acid provides more $H_3O^+$ ions. So the concentration of $H_3O^+$ (i.e., $ [H_3O^{+1}] $) increases.
# Answer: up (or increases)

### Sub - question 2
# Explanation:
## Step1: Recall pH formula
The formula for pH is $pH =-\log[H_3O^+]$.
## Step2: Analyze pH change with $ [H_3O^+] $
From sub - question 1, we know that $ [H_3O^+] $ increases when acid is added. Since the logarithm function $y =-\log(x)$ is a decreasing function (as $x$ increases, $y$ decreases), when $ [H_3O^+] $ increases, the pH value (which is $ -\log[H_3O^+] $) decreases. For example, if $ [H_3O^+]=1\times10^{-7}$ (neutral water), $pH = 7$; if $ [H_3O^+]=1\times10^{-6}$, $pH=-\log(1\times10^{-6}) = 6$, which is a decrease from 7.
# Answer: down (or decreases)

### Sub - question 3
# Explanation:
## Step1: Recall $K_w$ expression
We know that $K_w=[H_3O^+][OH^-]=1\times10^{-14}$ (at $25^{\circ}C$).
## Step2: Analyze $ [OH^{-1}] $ change
From sub - question 1, $ [H_3O^+] $ increases. Since $K_w$ is a constant (at a given temperature), if $ [H_3O^+] $ increases, then $ [OH^-]=\frac{K_w}{[H_3O^+]}$ must decrease (because $K_w$ is fixed and the numerator is constant while the denominator increases).
# Answer: down (or decreases)

### Sub - question 4
# Explanation:
## Step1: Recall pOH formula
The formula for pOH is $pOH =-\log[OH^-]$.
## Step2: Analyze pOH change with $ [OH^-] $
From sub - question 3, $ [OH^-] $ decreases when acid is added. Since $y =-\log(x)$ is a decreasing function, when $x$ (here $x = [OH^-]$) decreases, $y$ (pOH) increases. For example, if $ [OH^-]=1\times10^{-7}$ (neutral water), $pOH = 7$; if $ [OH^-]=1\times10^{-8}$, $pOH=-\log(1\times10^{-8}) = 8$, which is an increase from 7. Also, from $pH + pOH=14$, when pH decreases (from sub - question 2), pOH must increase.
# Answer: up (or increases)

### Sub - question 5
# Explanation:
## Step1: Recall $K_w$ definition
$K_w=[H_3O^+][OH^-]$ and $K_w$ is a constant at a given temperature (like $25^{\circ}C$ where $K_w = 1\times10^{-14}$).
## Step2: Analyze $K_w$ change with acid addition
Adding acid to water does not change the temperature (assuming the process is isothermal). Since $K_w$ depends only on temperature, at constant temperature, $K_w$ remains the same even when acid is added to neutral water.
# Answer: remains the same (or no change)

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QUESTION IMAGE

Question

Explanation:

Sub - question 1

Step1: Recall acid - water reaction

Step2: Analyze \( [H_3O^{+1}] \) change

Step1: Recall pH formula

Step2: Analyze pH change with \( [H_3O^+] \)

Step1: Recall \(K_w\) expression

Step2: Analyze \( [OH^{-1}] \) change

Answer:

Sub - question 2