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Sagot :
To determine the concentration of hydronium ions ([tex]\(\left[ H_3O^+ \right]\)[/tex]) in lemon juice with a given pH of 2.32, we can use the formula:
[tex]\[ \left[ H_3O^+ \right] = 10^{-\text{pH}} \][/tex]
Given:
[tex]\[ \text{pH} = 2.32 \][/tex]
We substitute the pH value into the formula to find the concentration:
[tex]\[ \left[ H_3O^+ \right] = 10^{-2.32} \][/tex]
Evaluating [tex]\(10^{-2.32}\)[/tex]:
[tex]\[ 10^{-2.32} \approx 0.004786 \][/tex]
Therefore, the concentration of [tex]\(H_3O^+\)[/tex] ions in the lemon juice is approximately [tex]\(0.004786\)[/tex] M.
Among the given options:
- [tex]\(1.05 \times 10^{-3} M\)[/tex]
- [tex]\(4.79 \times 10^{-3} M\)[/tex]
- [tex]\(2.08 \times 10^2 M\)[/tex]
- [tex]\(9.55 \times 10^2 M\)[/tex]
The concentration [tex]\(0.004786\)[/tex] M matches most closely with [tex]\( 4.79 \times 10^{-3} M \)[/tex].
Thus, the correct answer is:
[tex]\[ 4.79 \times 10^{-3} M \][/tex]
[tex]\[ \left[ H_3O^+ \right] = 10^{-\text{pH}} \][/tex]
Given:
[tex]\[ \text{pH} = 2.32 \][/tex]
We substitute the pH value into the formula to find the concentration:
[tex]\[ \left[ H_3O^+ \right] = 10^{-2.32} \][/tex]
Evaluating [tex]\(10^{-2.32}\)[/tex]:
[tex]\[ 10^{-2.32} \approx 0.004786 \][/tex]
Therefore, the concentration of [tex]\(H_3O^+\)[/tex] ions in the lemon juice is approximately [tex]\(0.004786\)[/tex] M.
Among the given options:
- [tex]\(1.05 \times 10^{-3} M\)[/tex]
- [tex]\(4.79 \times 10^{-3} M\)[/tex]
- [tex]\(2.08 \times 10^2 M\)[/tex]
- [tex]\(9.55 \times 10^2 M\)[/tex]
The concentration [tex]\(0.004786\)[/tex] M matches most closely with [tex]\( 4.79 \times 10^{-3} M \)[/tex].
Thus, the correct answer is:
[tex]\[ 4.79 \times 10^{-3} M \][/tex]
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