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Sagot :
Certainly! Let's go through each part of the question step-by-step to find the concentration of hydronium ions, [tex]\( \text{H}_3\text{O}^+ \)[/tex], in different scenarios.
### Part a: Given [tex]\([ \text{OH}^- ] = 5 \times 10^{-5} \text{ M}\)[/tex]
We use the ion-product constant for water at 25 degrees Celsius:
[tex]\[ K_w = [\text{H}_3\text{O}^+][\text{OH}^-] = 1 \times 10^{-14} \][/tex]
We can find the concentration of [tex]\(\text{H}_3\text{O}^+\)[/tex] using the given [tex]\([ \text{OH}^- ]\)[/tex]:
[tex]\[ [\text{H}_3\text{O}^+] = \frac{K_w}{[ \text{OH}^- ]} \][/tex]
Substitute the values:
[tex]\[ [\text{H}_3\text{O}^+] = \frac{1 \times 10^{-14}}{5 \times 10^{-5}} \][/tex]
[tex]\[ [\text{H}_3\text{O}^+] = 2 \times 10^{-10} \text{ M} \][/tex]
So, the concentration of hydronium ions is [tex]\( 2 \times 10^{-10} \text{ M} \)[/tex].
### Part c: Given [tex]\(0.05 \text{ M Sr(OH)}_2\)[/tex]
Strontium hydroxide, [tex]\( \text{Sr(OH)}_2 \)[/tex], dissociates completely in water and provides 2 hydroxide ions ([tex]\( \text{OH}^- \)[/tex]) per formula unit:
[tex]\[ \text{Sr(OH)}_2 \rightarrow \text{Sr}^{2+} + 2\text{OH}^- \][/tex]
Therefore, the concentration of hydroxide ions is:
[tex]\[ [ \text{OH}^- ] = 2 \times 0.05 \text{ M} = 0.10 \text{ M} \][/tex]
Now, we use the ion-product constant for water:
[tex]\[ [\text{H}_3\text{O}^+] = \frac{K_w}{[ \text{OH}^- ]} \][/tex]
Substitute the values:
[tex]\[ [\text{H}_3\text{O}^+] = \frac{1 \times 10^{-14}}{0.10} \][/tex]
[tex]\[ [\text{H}_3\text{O}^+] = 1 \times 10^{-13} \text{ M} \][/tex]
So, the concentration of hydronium ions is [tex]\( 1 \times 10^{-13} \text{ M} \)[/tex].
### Part b: Given [tex]\([ \text{OH}^- ] = 100 [ \text{H}_3\text{O}^+ ]\)[/tex]
This gives us a relationship between [tex]\([ \text{OH}^- ]\)[/tex] and [tex]\([ \text{H}_3\text{O}^+ ]\)[/tex]:
[tex]\[ [ \text{OH}^- ] = 100 [ \text{H}_3\text{O}^+ ] \][/tex]
We use the ion-product constant for water:
[tex]\[ K_w = [\text{H}_3\text{O}^+][\text{OH}^-] \][/tex]
Using the given relationship:
[tex]\[ K_w = [\text{H}_3\text{O}^+] \cdot 100 [ \text{H}_3\text{O}^+ ] \][/tex]
[tex]\[ K_w = 100 [\text{H}_3\text{O}^+]^2 \][/tex]
Solving for [tex]\([ \text{H}_3\text{O}^+] \)[/tex]:
[tex]\[ [\text{H}_3\text{O}^+]^2 = \frac{K_w}{100} \][/tex]
[tex]\[ [\text{H}_3\text{O}^+] = \sqrt{\frac{1 \times 10^{-14}}{100}} = \sqrt{1 \times 10^{-16}} \][/tex]
[tex]\[ [\text{H}_3\text{O}^+] = 1 \times 10^{-8} \text{ M} \][/tex]
So, the concentration of hydronium ions is [tex]\( 1 \times 10^{-8} \text{ M} \)[/tex].
### Part d: Given [tex]\(0.01 \text{ M H}_2\text{SO}_4\)[/tex]
Sulfuric acid, [tex]\( \text{H}_2\text{SO}_4 \)[/tex], is a strong acid that dissociates completely in water and provides 2 hydronium ions ([tex]\( \text{H}_3\text{O}^+ \)[/tex]) per formula unit:
[tex]\[ \text{H}_2\text{SO}_4 \rightarrow 2\text{H}_3\text{O}^+ + \text{SO}_4^{2-} \][/tex]
Therefore, the concentration of hydronium ions is:
[tex]\[ [\text{H}_3\text{O}^+] = 2 \times 0.01 \text{ M} = 0.02 \text{ M} \][/tex]
So, the concentration of hydronium ions is [tex]\( 0.02 \text{ M} \)[/tex].
### Summary:
a. [tex]\( [\text{H}_3\text{O}^+] = 2 \times 10^{-10} \text{ M} \)[/tex]
c. [tex]\( [\text{H}_3\text{O}^+] = 1 \times 10^{-13} \text{ M} \)[/tex]
b. [tex]\( [\text{H}_3\text{O}^+] = 1 \times 10^{-8} \text{ M} \)[/tex]
d. [tex]\( [\text{H}_3\text{O}^+] = 0.02 \text{ M} \)[/tex]
### Part a: Given [tex]\([ \text{OH}^- ] = 5 \times 10^{-5} \text{ M}\)[/tex]
We use the ion-product constant for water at 25 degrees Celsius:
[tex]\[ K_w = [\text{H}_3\text{O}^+][\text{OH}^-] = 1 \times 10^{-14} \][/tex]
We can find the concentration of [tex]\(\text{H}_3\text{O}^+\)[/tex] using the given [tex]\([ \text{OH}^- ]\)[/tex]:
[tex]\[ [\text{H}_3\text{O}^+] = \frac{K_w}{[ \text{OH}^- ]} \][/tex]
Substitute the values:
[tex]\[ [\text{H}_3\text{O}^+] = \frac{1 \times 10^{-14}}{5 \times 10^{-5}} \][/tex]
[tex]\[ [\text{H}_3\text{O}^+] = 2 \times 10^{-10} \text{ M} \][/tex]
So, the concentration of hydronium ions is [tex]\( 2 \times 10^{-10} \text{ M} \)[/tex].
### Part c: Given [tex]\(0.05 \text{ M Sr(OH)}_2\)[/tex]
Strontium hydroxide, [tex]\( \text{Sr(OH)}_2 \)[/tex], dissociates completely in water and provides 2 hydroxide ions ([tex]\( \text{OH}^- \)[/tex]) per formula unit:
[tex]\[ \text{Sr(OH)}_2 \rightarrow \text{Sr}^{2+} + 2\text{OH}^- \][/tex]
Therefore, the concentration of hydroxide ions is:
[tex]\[ [ \text{OH}^- ] = 2 \times 0.05 \text{ M} = 0.10 \text{ M} \][/tex]
Now, we use the ion-product constant for water:
[tex]\[ [\text{H}_3\text{O}^+] = \frac{K_w}{[ \text{OH}^- ]} \][/tex]
Substitute the values:
[tex]\[ [\text{H}_3\text{O}^+] = \frac{1 \times 10^{-14}}{0.10} \][/tex]
[tex]\[ [\text{H}_3\text{O}^+] = 1 \times 10^{-13} \text{ M} \][/tex]
So, the concentration of hydronium ions is [tex]\( 1 \times 10^{-13} \text{ M} \)[/tex].
### Part b: Given [tex]\([ \text{OH}^- ] = 100 [ \text{H}_3\text{O}^+ ]\)[/tex]
This gives us a relationship between [tex]\([ \text{OH}^- ]\)[/tex] and [tex]\([ \text{H}_3\text{O}^+ ]\)[/tex]:
[tex]\[ [ \text{OH}^- ] = 100 [ \text{H}_3\text{O}^+ ] \][/tex]
We use the ion-product constant for water:
[tex]\[ K_w = [\text{H}_3\text{O}^+][\text{OH}^-] \][/tex]
Using the given relationship:
[tex]\[ K_w = [\text{H}_3\text{O}^+] \cdot 100 [ \text{H}_3\text{O}^+ ] \][/tex]
[tex]\[ K_w = 100 [\text{H}_3\text{O}^+]^2 \][/tex]
Solving for [tex]\([ \text{H}_3\text{O}^+] \)[/tex]:
[tex]\[ [\text{H}_3\text{O}^+]^2 = \frac{K_w}{100} \][/tex]
[tex]\[ [\text{H}_3\text{O}^+] = \sqrt{\frac{1 \times 10^{-14}}{100}} = \sqrt{1 \times 10^{-16}} \][/tex]
[tex]\[ [\text{H}_3\text{O}^+] = 1 \times 10^{-8} \text{ M} \][/tex]
So, the concentration of hydronium ions is [tex]\( 1 \times 10^{-8} \text{ M} \)[/tex].
### Part d: Given [tex]\(0.01 \text{ M H}_2\text{SO}_4\)[/tex]
Sulfuric acid, [tex]\( \text{H}_2\text{SO}_4 \)[/tex], is a strong acid that dissociates completely in water and provides 2 hydronium ions ([tex]\( \text{H}_3\text{O}^+ \)[/tex]) per formula unit:
[tex]\[ \text{H}_2\text{SO}_4 \rightarrow 2\text{H}_3\text{O}^+ + \text{SO}_4^{2-} \][/tex]
Therefore, the concentration of hydronium ions is:
[tex]\[ [\text{H}_3\text{O}^+] = 2 \times 0.01 \text{ M} = 0.02 \text{ M} \][/tex]
So, the concentration of hydronium ions is [tex]\( 0.02 \text{ M} \)[/tex].
### Summary:
a. [tex]\( [\text{H}_3\text{O}^+] = 2 \times 10^{-10} \text{ M} \)[/tex]
c. [tex]\( [\text{H}_3\text{O}^+] = 1 \times 10^{-13} \text{ M} \)[/tex]
b. [tex]\( [\text{H}_3\text{O}^+] = 1 \times 10^{-8} \text{ M} \)[/tex]
d. [tex]\( [\text{H}_3\text{O}^+] = 0.02 \text{ M} \)[/tex]
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