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Sagot :
Certainly! Let's walk through the solutions step-by-step for each part of the question.
### 47. Calculate the [tex]\([H_3O^+]\)[/tex] of a solution:
#### (a) [tex]\( pH = 5 \)[/tex]
To find the hydronium ion concentration [tex]\([H_3O^+]\)[/tex], we use the formula:
[tex]\[ [H_3O^+] = 10^{-pH} \][/tex]
For [tex]\( pH = 5 \)[/tex]:
[tex]\[ [H_3O^+] = 10^{-5} = 1 \times 10^{-5} \, M \][/tex]
#### (b) [tex]\( pH = 3.2 \)[/tex]
Again, using the formula:
[tex]\[ [H_3O^+] = 10^{-pH} \][/tex]
For [tex]\( pH = 3.2 \)[/tex]:
[tex]\[ [H_3O^+] \approx 10^{-3.2} \approx 0.000630957344480193 \, M \][/tex]
#### (c) [tex]\( pOH = 9.3 \)[/tex]
First, we need to convert [tex]\( pOH \)[/tex] to [tex]\( pH \)[/tex] using the relationship:
[tex]\[ pH + pOH = 14 \][/tex]
Thus,
[tex]\[ pH = 14 - pOH = 14 - 9.3 = 4.7 \][/tex]
Then, using the formula:
[tex]\[ [H_3O^+] = 10^{-pH} \][/tex]
[tex]\[ [H_3O^+] \approx 10^{-4.7} \approx 5.011872336272714 \times 10^{-10} \, M \][/tex]
#### (d) [tex]\( pOH = 11.3 \)[/tex]
Similarly, convert [tex]\( pOH \)[/tex] to [tex]\( pH \)[/tex]:
[tex]\[ pH = 14 - pOH = 14 - 11.3 = 2.7 \][/tex]
Then, using the formula:
[tex]\[ [H_3O^+] = 10^{-pH} \][/tex]
[tex]\[ [H_3O^+] \approx 10^{-2.7} \approx 5.011872336272715 \times 10^{-12} \, M \][/tex]
### 48. Calculate the [tex]\([OH^-]\)[/tex] or [tex]\([H_3O^+]\)[/tex] of a solution given the following. (Remember [tex]\( K_w = 1 \times 10^{-14} \, M^2 \)[/tex])
#### (a) [tex]\([H_3O^+] = 1 \times 10^{-6} \, M\)[/tex]
Using the water dissociation constant [tex]\( K_w = [H_3O^+][OH^-] \)[/tex]:
[tex]\[ [OH^-] = \frac{K_w}{[H_3O^+]} = \frac{1 \times 10^{-14}}{1 \times 10^{-6}} = 1 \times 10^{-8} \, M \][/tex]
#### (b) [tex]\([H_3O^+] = 1 \times 10^{-1} \, M\)[/tex]
Similarly, using the same relationship:
[tex]\[ [OH^-] = \frac{K_w}{[H_3O^+]} = \frac{1 \times 10^{-14}}{1 \times 10^{-1}} = 1 \times 10^{-13} \, M \][/tex]
#### (c) [tex]\([OH^-] = 1 \times 10^{-12} \, M\)[/tex]
Again, using [tex]\( K_w \)[/tex]:
[tex]\[ [H_3O^+] = \frac{K_w}{[OH^-]} = \frac{1 \times 10^{-14}}{1 \times 10^{-12}} = 1 \times 10^{-2} \, M \][/tex]
### 49. Will a high or low pH result from the following?
#### (a) Low [tex]\([H_3O^+]\)[/tex]
A low concentration of [tex]\([H_3O^+]\)[/tex] means the solution is less acidic, which results in a high pH. Thus,
[tex]\[ \text{High pH} \][/tex]
#### (b) High [tex]\([H_3O^+]\)[/tex]
A high concentration of [tex]\([H_3O^+]\)[/tex] means the solution is more acidic, which results in a low pH. Thus,
[tex]\[ \text{Low pH} \][/tex]
#### (c) Low [tex]\([OH^-]\)[/tex]
A low concentration of [tex]\([OH^-]\)[/tex] implies a higher concentration of [tex]\([H_3O^+]\)[/tex] (as [tex]\( K_w \)[/tex] is constant), leading to a more acidic solution with a low pH. Thus,
[tex]\[ \text{Low pH} \][/tex]
#### (d) High [tex]\([OH^-]\)[/tex]
A high concentration of [tex]\([OH^-]\)[/tex] implies a lower concentration of [tex]\([H_3O^+]\)[/tex], leading to a more basic solution with a high pH. Thus,
[tex]\[ \text{High pH} \][/tex]
These results conclude the step-by-step solution for each part of the question.
### 47. Calculate the [tex]\([H_3O^+]\)[/tex] of a solution:
#### (a) [tex]\( pH = 5 \)[/tex]
To find the hydronium ion concentration [tex]\([H_3O^+]\)[/tex], we use the formula:
[tex]\[ [H_3O^+] = 10^{-pH} \][/tex]
For [tex]\( pH = 5 \)[/tex]:
[tex]\[ [H_3O^+] = 10^{-5} = 1 \times 10^{-5} \, M \][/tex]
#### (b) [tex]\( pH = 3.2 \)[/tex]
Again, using the formula:
[tex]\[ [H_3O^+] = 10^{-pH} \][/tex]
For [tex]\( pH = 3.2 \)[/tex]:
[tex]\[ [H_3O^+] \approx 10^{-3.2} \approx 0.000630957344480193 \, M \][/tex]
#### (c) [tex]\( pOH = 9.3 \)[/tex]
First, we need to convert [tex]\( pOH \)[/tex] to [tex]\( pH \)[/tex] using the relationship:
[tex]\[ pH + pOH = 14 \][/tex]
Thus,
[tex]\[ pH = 14 - pOH = 14 - 9.3 = 4.7 \][/tex]
Then, using the formula:
[tex]\[ [H_3O^+] = 10^{-pH} \][/tex]
[tex]\[ [H_3O^+] \approx 10^{-4.7} \approx 5.011872336272714 \times 10^{-10} \, M \][/tex]
#### (d) [tex]\( pOH = 11.3 \)[/tex]
Similarly, convert [tex]\( pOH \)[/tex] to [tex]\( pH \)[/tex]:
[tex]\[ pH = 14 - pOH = 14 - 11.3 = 2.7 \][/tex]
Then, using the formula:
[tex]\[ [H_3O^+] = 10^{-pH} \][/tex]
[tex]\[ [H_3O^+] \approx 10^{-2.7} \approx 5.011872336272715 \times 10^{-12} \, M \][/tex]
### 48. Calculate the [tex]\([OH^-]\)[/tex] or [tex]\([H_3O^+]\)[/tex] of a solution given the following. (Remember [tex]\( K_w = 1 \times 10^{-14} \, M^2 \)[/tex])
#### (a) [tex]\([H_3O^+] = 1 \times 10^{-6} \, M\)[/tex]
Using the water dissociation constant [tex]\( K_w = [H_3O^+][OH^-] \)[/tex]:
[tex]\[ [OH^-] = \frac{K_w}{[H_3O^+]} = \frac{1 \times 10^{-14}}{1 \times 10^{-6}} = 1 \times 10^{-8} \, M \][/tex]
#### (b) [tex]\([H_3O^+] = 1 \times 10^{-1} \, M\)[/tex]
Similarly, using the same relationship:
[tex]\[ [OH^-] = \frac{K_w}{[H_3O^+]} = \frac{1 \times 10^{-14}}{1 \times 10^{-1}} = 1 \times 10^{-13} \, M \][/tex]
#### (c) [tex]\([OH^-] = 1 \times 10^{-12} \, M\)[/tex]
Again, using [tex]\( K_w \)[/tex]:
[tex]\[ [H_3O^+] = \frac{K_w}{[OH^-]} = \frac{1 \times 10^{-14}}{1 \times 10^{-12}} = 1 \times 10^{-2} \, M \][/tex]
### 49. Will a high or low pH result from the following?
#### (a) Low [tex]\([H_3O^+]\)[/tex]
A low concentration of [tex]\([H_3O^+]\)[/tex] means the solution is less acidic, which results in a high pH. Thus,
[tex]\[ \text{High pH} \][/tex]
#### (b) High [tex]\([H_3O^+]\)[/tex]
A high concentration of [tex]\([H_3O^+]\)[/tex] means the solution is more acidic, which results in a low pH. Thus,
[tex]\[ \text{Low pH} \][/tex]
#### (c) Low [tex]\([OH^-]\)[/tex]
A low concentration of [tex]\([OH^-]\)[/tex] implies a higher concentration of [tex]\([H_3O^+]\)[/tex] (as [tex]\( K_w \)[/tex] is constant), leading to a more acidic solution with a low pH. Thus,
[tex]\[ \text{Low pH} \][/tex]
#### (d) High [tex]\([OH^-]\)[/tex]
A high concentration of [tex]\([OH^-]\)[/tex] implies a lower concentration of [tex]\([H_3O^+]\)[/tex], leading to a more basic solution with a high pH. Thus,
[tex]\[ \text{High pH} \][/tex]
These results conclude the step-by-step solution for each part of the question.
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