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Sagot :
To solve this problem, we need to compare the hydronium ion concentrations [tex]\([H_3O^+]\)[/tex] for the various given measurements and rank them from highest to lowest.
Given:
1. [tex]\([H_3O^+] = 3.16 \times 10^{-4} \, M\)[/tex]
2. [tex]\([HH^2 = 4.35 \times 10^2 \, N]\)[/tex] (This value is irrelevant in this context)
3. [tex]\( pH = 1.05 \)[/tex]
4. [tex]\( pOH = 7.0 \)[/tex]
5. [tex]\( pOH = 4.0 \)[/tex]
### Step-by-step Solution:
#### Step 1: Calculate [tex]\([H_3O^+]\)[/tex] from [tex]\( pH \)[/tex]
For [tex]\( pH = 1.05 \)[/tex]:
The relationship between [tex]\([H_3O^+]\)[/tex] and [tex]\( pH \)[/tex] is given by:
[tex]\[ [H_3O^+] = 10^{-pH} \][/tex]
So,
[tex]\[ [H_3O^+] = 10^{-1.05} \approx 0.0891 \][/tex]
#### Step 2: Calculate [tex]\([H_3O^+]\)[/tex] from [tex]\( pOH = 7.0 \)[/tex]
Using the relationship [tex]\( pH + pOH = 14 \)[/tex]:
[tex]\[ pH = 14 - pOH = 14 - 7.0 = 7.0 \][/tex]
Then, using the same formula:
[tex]\[ [H_3O^+] = 10^{-pH} \][/tex]
[tex]\[ [H_3O^+] = 10^{-7.0} = 10^{-7} = 1.0 \times 10^{-7} \, M \][/tex]
#### Step 3: Calculate [tex]\([H_3O^+]\)[/tex] from [tex]\( pOH = 4.0 \)[/tex]
Using the relationship [tex]\( pH + pOH = 14 \)[/tex]:
[tex]\[ pH = 14 - pOH = 14 - 4.0 = 10.0 \][/tex]
Then, using the same formula:
[tex]\[ [H_3O^+] = 10^{-pH} \][/tex]
[tex]\[ [H_3O^+] = 10^{-10.0} = 10^{-10} = 1.0 \times 10^{-10} \, M \][/tex]
### Summary of Calculated [tex]\([H_3O^+]\)[/tex] Values:
1. Given: [tex]\( 3.16 \times 10^{-4} \, M \)[/tex]
2. Calculated from [tex]\( pH = 1.05 \)[/tex]: [tex]\( 0.0891 \, M \)[/tex]
3. Calculated from [tex]\( pOH = 7.0 \)[/tex]: [tex]\( 1.0 \times 10^{-7} \, M \)[/tex]
4. Calculated from [tex]\( pOH = 4.0 \)[/tex]: [tex]\( 1.0 \times 10^{-10} \, M \)[/tex]
### Step 4: Rank these values from highest to lowest:
- [tex]\( 0.0891 \, M \)[/tex] (derived from [tex]\( pH = 1.05 \)[/tex])
- [tex]\( 3.16 \times 10^{-4} \, M \)[/tex]
- [tex]\( 1.0 \times 10^{-7} \, M \)[/tex] (derived from [tex]\( pOH = 7.0 \)[/tex])
- [tex]\( 1.0 \times 10^{-10} \, M \)[/tex] (derived from [tex]\( pOH = 4.0 \)[/tex])
### Final Ranking:
1. [tex]\( 0.0891 \, M \)[/tex] (Solution derived from [tex]\( pH = 1.05 \)[/tex])
2. [tex]\( 3.16 \times 10^{-4} \, M \)[/tex] (Given concentration)
3. [tex]\( 1.0 \times 10^{-7} \, M \)[/tex] (Solution derived from [tex]\( pOH = 7.0 \)[/tex])
4. [tex]\( 1.0 \times 10^{-10} \, M \)[/tex] (Solution derived from [tex]\( pOH = 4.0 \)[/tex])
Therefore, the order is:
- For [tex]\(\left[ H _3 O ^{+}\right] = 0.0891 \, M\)[/tex], rank 1
- For [tex]\(\left[ H _3 O ^{+}\right] = 3.16 \times 10^{-4} \, M\)[/tex], rank 2
- For [tex]\(\left[ H _3 O ^{+}\right] = 1.0 \times 10^{-7} \, M\)[/tex], rank 3
- For [tex]\(\left[ H _3 O ^{+}\right] = 1.0 \times 10^{-10} \, M\)[/tex], rank 4
Given:
1. [tex]\([H_3O^+] = 3.16 \times 10^{-4} \, M\)[/tex]
2. [tex]\([HH^2 = 4.35 \times 10^2 \, N]\)[/tex] (This value is irrelevant in this context)
3. [tex]\( pH = 1.05 \)[/tex]
4. [tex]\( pOH = 7.0 \)[/tex]
5. [tex]\( pOH = 4.0 \)[/tex]
### Step-by-step Solution:
#### Step 1: Calculate [tex]\([H_3O^+]\)[/tex] from [tex]\( pH \)[/tex]
For [tex]\( pH = 1.05 \)[/tex]:
The relationship between [tex]\([H_3O^+]\)[/tex] and [tex]\( pH \)[/tex] is given by:
[tex]\[ [H_3O^+] = 10^{-pH} \][/tex]
So,
[tex]\[ [H_3O^+] = 10^{-1.05} \approx 0.0891 \][/tex]
#### Step 2: Calculate [tex]\([H_3O^+]\)[/tex] from [tex]\( pOH = 7.0 \)[/tex]
Using the relationship [tex]\( pH + pOH = 14 \)[/tex]:
[tex]\[ pH = 14 - pOH = 14 - 7.0 = 7.0 \][/tex]
Then, using the same formula:
[tex]\[ [H_3O^+] = 10^{-pH} \][/tex]
[tex]\[ [H_3O^+] = 10^{-7.0} = 10^{-7} = 1.0 \times 10^{-7} \, M \][/tex]
#### Step 3: Calculate [tex]\([H_3O^+]\)[/tex] from [tex]\( pOH = 4.0 \)[/tex]
Using the relationship [tex]\( pH + pOH = 14 \)[/tex]:
[tex]\[ pH = 14 - pOH = 14 - 4.0 = 10.0 \][/tex]
Then, using the same formula:
[tex]\[ [H_3O^+] = 10^{-pH} \][/tex]
[tex]\[ [H_3O^+] = 10^{-10.0} = 10^{-10} = 1.0 \times 10^{-10} \, M \][/tex]
### Summary of Calculated [tex]\([H_3O^+]\)[/tex] Values:
1. Given: [tex]\( 3.16 \times 10^{-4} \, M \)[/tex]
2. Calculated from [tex]\( pH = 1.05 \)[/tex]: [tex]\( 0.0891 \, M \)[/tex]
3. Calculated from [tex]\( pOH = 7.0 \)[/tex]: [tex]\( 1.0 \times 10^{-7} \, M \)[/tex]
4. Calculated from [tex]\( pOH = 4.0 \)[/tex]: [tex]\( 1.0 \times 10^{-10} \, M \)[/tex]
### Step 4: Rank these values from highest to lowest:
- [tex]\( 0.0891 \, M \)[/tex] (derived from [tex]\( pH = 1.05 \)[/tex])
- [tex]\( 3.16 \times 10^{-4} \, M \)[/tex]
- [tex]\( 1.0 \times 10^{-7} \, M \)[/tex] (derived from [tex]\( pOH = 7.0 \)[/tex])
- [tex]\( 1.0 \times 10^{-10} \, M \)[/tex] (derived from [tex]\( pOH = 4.0 \)[/tex])
### Final Ranking:
1. [tex]\( 0.0891 \, M \)[/tex] (Solution derived from [tex]\( pH = 1.05 \)[/tex])
2. [tex]\( 3.16 \times 10^{-4} \, M \)[/tex] (Given concentration)
3. [tex]\( 1.0 \times 10^{-7} \, M \)[/tex] (Solution derived from [tex]\( pOH = 7.0 \)[/tex])
4. [tex]\( 1.0 \times 10^{-10} \, M \)[/tex] (Solution derived from [tex]\( pOH = 4.0 \)[/tex])
Therefore, the order is:
- For [tex]\(\left[ H _3 O ^{+}\right] = 0.0891 \, M\)[/tex], rank 1
- For [tex]\(\left[ H _3 O ^{+}\right] = 3.16 \times 10^{-4} \, M\)[/tex], rank 2
- For [tex]\(\left[ H _3 O ^{+}\right] = 1.0 \times 10^{-7} \, M\)[/tex], rank 3
- For [tex]\(\left[ H _3 O ^{+}\right] = 1.0 \times 10^{-10} \, M\)[/tex], rank 4
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