What is the [OH^-] for a solution at 25°C that has [H₃O⁺] = 2.35 $\times$ 10^-3 M?

Question

Quizplus · Accepted Answer

Using the formula for calculating [OH-] from [H+], we have:
pH + pOH = 14
pH = -log[H+] = -log(2.35x10^-3) = 2.63
pOH = 14 - pH = 11.37
Using the formula for calculating [OH-] from pOH, we have:
pOH = -log[OH-]
11.37 = -log[OH-]
[OH-] = 4.26x10^-12 M
Now, we can use the equation for K w to calculate [H+] and [OH-]:
K w = [H+][OH-] = 1.0x10^-14
1.0x10^-14 = [H+][4.26x10^-12]
[H+] = 2.35x10^-3 M
Therefore, [OHS1U1P1-S1S1P0] = [OH-] = 4.26x10^-12 M
[HS1U1B13S1U1B0OS1U1P1+S1S1P0] = [H+] + [HSO4-] = 2.35x10^-3 M (assuming the solution only contains H2SO4 and H2O)
Using the equation for the dissociation of H2SO4, we know that:
[H+] = 2[HSO4-]
[HSO4-] = [H+] / 2 = 2.35x10^-3 / 2 = 1.175x10^-3 M
Therefore, [HS1U1B13S1U1B0OS1U1P1] = [HSO4-] = 1.175x10^-3 M
Adding [OHS1U1P1-S1S1P0] to both sides, we get:
[HS1U1B13S1U1B0OS1U1P1+S1S1P0] + [OHS1U1P1-S1S1P0] = 2.35x10^-3 M + 4.26x10^-12 M = 2.35 F1F1F1S1F1F1F10 10S1U1P1-3S1S1P0 M 
Therefore, the answer is choice C.

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