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Sagot :
To determine which solution is the most acidic, we need to identify which one has the highest concentration of hydronium ions [tex]\([H_3O^+]\)[/tex]. The higher the concentration of hydronium ions, the more acidic the solution is.
We are given four different concentrations:
- Two concentrations of hydroxide ions [tex]\([OH^-]\)[/tex]:
1. [tex]\([OH^-] = 2 \times 10^{-12} \text{M}\)[/tex]
2. [tex]\([OH^-] = 3 \times 10^{-3} \text{M}\)[/tex]
- Two concentrations of hydronium ions [tex]\([H_3O^+]\)[/tex]:
1. [tex]\([H_3O^+] = 4 \times 10^{-4} \text{M}\)[/tex]
2. [tex]\([H_3O^+] = 6 \times 10^{-6} \text{M}\)[/tex]
For the given [tex]\([OH^-]\)[/tex] concentrations, we need to convert them to [tex]\([H_3O^+]\)[/tex] using the relationship:
[tex]\[ [H_3O^+] \times [OH^-] = 1 \times 10^{-14} \][/tex]
Let's calculate the [tex]\([H_3O^+]\)[/tex] concentrations:
1. For [tex]\([OH^-] = 2 \times 10^{-12} \text{M}\)[/tex]:
[tex]\[ [H_3O^+] = \frac{1 \times 10^{-14}}{2 \times 10^{-12}} = \frac{1 \times 10^{-14}}{2 \times 10^{-12}} = 0.005 \text{M} \][/tex]
2. For [tex]\([OH^-] = 3 \times 10^{-3} \text{M}\)[/tex]:
[tex]\[ [H_3O^+] = \frac{1 \times 10^{-14}}{3 \times 10^{-3}} = \frac{1 \times 10^{-14}}{0.003} \approx 3.333 \times 10^{-12} \text{M} \][/tex]
Now, we have four [tex]\([H_3O^+]\)[/tex] concentrations:
1. From [tex]\([OH^-] = 2 \times 10^{-12} \text{M}\)[/tex]: [tex]\([H_3O^+] = 0.005 \text{M}\)[/tex]
2. From [tex]\([OH^-] = 3 \times 10^{-3} \text{M}\)[/tex]: [tex]\([H_3O^+] \approx 3.333 \times 10^{-12} \text{M}\)[/tex]
3. Given directly: [tex]\([H_3O^+] = 4 \times 10^{-4} \text{M}\)[/tex]
4. Given directly: [tex]\([H_3O^+] = 6 \times 10^{-6} \text{M}\)[/tex]
To find the most acidic solution, we compare these [tex]\([H_3O^+]\)[/tex] concentrations:
- [tex]\(0.005 \text{M}\)[/tex]
- [tex]\(3.333 \times 10^{-12} \text{M}\)[/tex]
- [tex]\(4 \times 10^{-4} \text{M}\)[/tex]
- [tex]\(6 \times 10^{-6} \text{M}\)[/tex]
Clearly, the highest [tex]\([H_3O^+]\)[/tex] concentration is [tex]\(0.005 \text{M}\)[/tex].
Therefore, the solution with [tex]\([OH^-] = 2 \times 10^{-12} \text{M}\)[/tex] is the most acidic.
We are given four different concentrations:
- Two concentrations of hydroxide ions [tex]\([OH^-]\)[/tex]:
1. [tex]\([OH^-] = 2 \times 10^{-12} \text{M}\)[/tex]
2. [tex]\([OH^-] = 3 \times 10^{-3} \text{M}\)[/tex]
- Two concentrations of hydronium ions [tex]\([H_3O^+]\)[/tex]:
1. [tex]\([H_3O^+] = 4 \times 10^{-4} \text{M}\)[/tex]
2. [tex]\([H_3O^+] = 6 \times 10^{-6} \text{M}\)[/tex]
For the given [tex]\([OH^-]\)[/tex] concentrations, we need to convert them to [tex]\([H_3O^+]\)[/tex] using the relationship:
[tex]\[ [H_3O^+] \times [OH^-] = 1 \times 10^{-14} \][/tex]
Let's calculate the [tex]\([H_3O^+]\)[/tex] concentrations:
1. For [tex]\([OH^-] = 2 \times 10^{-12} \text{M}\)[/tex]:
[tex]\[ [H_3O^+] = \frac{1 \times 10^{-14}}{2 \times 10^{-12}} = \frac{1 \times 10^{-14}}{2 \times 10^{-12}} = 0.005 \text{M} \][/tex]
2. For [tex]\([OH^-] = 3 \times 10^{-3} \text{M}\)[/tex]:
[tex]\[ [H_3O^+] = \frac{1 \times 10^{-14}}{3 \times 10^{-3}} = \frac{1 \times 10^{-14}}{0.003} \approx 3.333 \times 10^{-12} \text{M} \][/tex]
Now, we have four [tex]\([H_3O^+]\)[/tex] concentrations:
1. From [tex]\([OH^-] = 2 \times 10^{-12} \text{M}\)[/tex]: [tex]\([H_3O^+] = 0.005 \text{M}\)[/tex]
2. From [tex]\([OH^-] = 3 \times 10^{-3} \text{M}\)[/tex]: [tex]\([H_3O^+] \approx 3.333 \times 10^{-12} \text{M}\)[/tex]
3. Given directly: [tex]\([H_3O^+] = 4 \times 10^{-4} \text{M}\)[/tex]
4. Given directly: [tex]\([H_3O^+] = 6 \times 10^{-6} \text{M}\)[/tex]
To find the most acidic solution, we compare these [tex]\([H_3O^+]\)[/tex] concentrations:
- [tex]\(0.005 \text{M}\)[/tex]
- [tex]\(3.333 \times 10^{-12} \text{M}\)[/tex]
- [tex]\(4 \times 10^{-4} \text{M}\)[/tex]
- [tex]\(6 \times 10^{-6} \text{M}\)[/tex]
Clearly, the highest [tex]\([H_3O^+]\)[/tex] concentration is [tex]\(0.005 \text{M}\)[/tex].
Therefore, the solution with [tex]\([OH^-] = 2 \times 10^{-12} \text{M}\)[/tex] is the most acidic.
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