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Sagot :
To determine the hydronium ion concentration [tex]\(\left[ \text{H}_3\text{O}^+ \right]\)[/tex] given the pH of a solution, we use the definition of pH, which is:
[tex]\[ \text{pH} = -\log \left( \left[ \text{H}_3\text{O}^+ \right] \right) \][/tex]
Given this equation, we can solve for [tex]\(\left[ \text{H}_3\text{O}^+ \right]\)[/tex] by rearranging it as follows:
[tex]\[ \left[ \text{H}_3\text{O}^+ \right] = 10^{-\text{pH}} \][/tex]
In this specific problem, the pH is given as 3.40. Plugging in the value of pH into the formula, we get:
[tex]\[ \left[ \text{H}_3\text{O}^+ \right] = 10^{-3.40} \][/tex]
Evaluating this expression yields:
[tex]\[ \left[ \text{H}_3\text{O}^+ \right] = 0.00039810717055349735\; \text{M} \][/tex]
To express this in scientific notation for easier comparison with the given options, we can rewrite it as:
[tex]\[ \left[ \text{H}_3\text{O}^+ \right] = 3.98 \times 10^{-4}\; \text{M} \][/tex]
Thus, the correct answer is:
[tex]\[ 3.98 \times 10^{-4}\; \text{M} \][/tex]
Therefore, the [tex]\(\left[ \text{H}_3\text{O}^+ \right]\)[/tex] for the solution is:
[tex]\[ \boxed{3.98 \times 10^{-4}\; \text{M}} \][/tex]
[tex]\[ \text{pH} = -\log \left( \left[ \text{H}_3\text{O}^+ \right] \right) \][/tex]
Given this equation, we can solve for [tex]\(\left[ \text{H}_3\text{O}^+ \right]\)[/tex] by rearranging it as follows:
[tex]\[ \left[ \text{H}_3\text{O}^+ \right] = 10^{-\text{pH}} \][/tex]
In this specific problem, the pH is given as 3.40. Plugging in the value of pH into the formula, we get:
[tex]\[ \left[ \text{H}_3\text{O}^+ \right] = 10^{-3.40} \][/tex]
Evaluating this expression yields:
[tex]\[ \left[ \text{H}_3\text{O}^+ \right] = 0.00039810717055349735\; \text{M} \][/tex]
To express this in scientific notation for easier comparison with the given options, we can rewrite it as:
[tex]\[ \left[ \text{H}_3\text{O}^+ \right] = 3.98 \times 10^{-4}\; \text{M} \][/tex]
Thus, the correct answer is:
[tex]\[ 3.98 \times 10^{-4}\; \text{M} \][/tex]
Therefore, the [tex]\(\left[ \text{H}_3\text{O}^+ \right]\)[/tex] for the solution is:
[tex]\[ \boxed{3.98 \times 10^{-4}\; \text{M}} \][/tex]
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