Let's solve the questions step-by-step, making sure we understand each part in detail.
### 1. Amount of [tex]\(OH^{-}\)[/tex] Ions in Solution with [tex]\(pH = 13\)[/tex]
#### Step 1: Calculate [tex]\(pOH\)[/tex]
For any aqueous solution at 25°C, the relationship between [tex]\(pH\)[/tex] and [tex]\(pOH\)[/tex] is given by:
[tex]\[pH + pOH = 14\][/tex]
Given that [tex]\(pH = 13\)[/tex], we can find [tex]\(pOH\)[/tex]:
[tex]\[pOH = 14 - pH\][/tex]
[tex]\[pOH = 14 - 13\][/tex]
[tex]\[pOH = 1\][/tex]
#### Step 2: Calculate [tex]\( [OH^-] \)[/tex]
To find the concentration of hydroxide ions ([tex]\([OH^-]\)[/tex]) in moles per liter, we use the formula:
[tex]\[ [OH^-] = 10^{-\text{pOH}} \][/tex]
Plugging in [tex]\( pOH = 1 \)[/tex]:
[tex]\[ [OH^-] = 10^{-1} = 0.1 \][/tex]
So, the solution contains [tex]\( 0.1 \)[/tex] moles of [tex]\( OH^- \)[/tex] ions per liter.
### 2. Amount of [tex]\(H^{+}\)[/tex] Ions in Solution with [tex]\(pH = 13\)[/tex]
To find the concentration of hydrogen ions ([tex]\([H^+]\)[/tex]), we use the formula:
[tex]\[ [H^+] = 10^{-\text{pH}} \][/tex]
Given [tex]\( pH = 13 \)[/tex]:
[tex]\[ [H^+] = 10^{-13} \][/tex]
So, the solution contains [tex]\( 1 \times 10^{-13} \)[/tex] moles of [tex]\( H^+ \)[/tex] ions per liter.
### 3. [tex]\( pH \)[/tex] of a Solution with [tex]\([H^+] = 1.0 \times 10^{-4}\)[/tex]
To find the pH of a solution, we use the formula:
[tex]\[ pH = -\log_{10} ([H^+]) \][/tex]
Given [tex]\([H^+] = 1.0 \times 10^{-4}\)[/tex]:
[tex]\[ pH = -\log_{10} (1.0 \times 10^{-4}) \][/tex]
Since [tex]\(\log_{10} (10^{-4}) = -4\)[/tex]:
[tex]\[ pH = -(-4) \][/tex]
[tex]\[ pH = 4 \][/tex]
### Summary:
1. The solution with [tex]\( pH = 13 \)[/tex] has approximately [tex]\( 0.1 \)[/tex] moles of [tex]\( OH^{-} \)[/tex] ions per liter.
2. The same solution has [tex]\( 1 \times 10^{-13} \)[/tex] moles of [tex]\( H^{+} \)[/tex] ions per liter.
3. A different solution with [tex]\([H^+] = 1.0 \times 10^{-4}\)[/tex] would have a [tex]\( pH = 4 \)[/tex].
