We need to calculate the pH of a 0.70 M solution of hypoiodous acid (HOI), given that the acid dissociation constant (\(K_a\)) for HOI is \(3.2 \times 10^{-11}\).
The dissociation of HOI in water can be represented as:
\[
\text{HOI} \rightleftharpoons \text{H}^+ + \text{OI}^-
\]
The equilibrium expression for this reaction is:
\[
K_a = \frac{[\text{H}^+][\text{OI}^-]}{[\text{HOI}]}
\]
Assume that initially, the concentration of \(\text{H}^+\) and \(\text{OI}^-\) is 0, and the concentration of \(\text{HOI}\) is 0.70 M. Let \(x\) be the change in concentration at equilibrium:
- \([\text{H}^+] = x\)
- \([\text{OI}^-] = x\)
- \([\text{HOI}] = 0.70 - x\)
Substitute the equilibrium concentrations into the \(K_a\) expression:
\[
3.2 \times 10^{-11} = \frac{x^2}{0.70 - x}
\]
Since \(K_a\) is very small, we can assume \(x\) is much smaller than 0.70, so \(0.70 - x \approx 0.70\). This simplifies the equation to:
\[
3.2 \times 10^{-11} = \frac{x^2}{0.70}
\]
Solve for \(x\) by multiplying both sides by 0.70:
\[
x^2 = 3.2 \times 10^{-11} \times 0.70
\]
\[
x^2 = 2.24 \times 10^{-11}
\]
\[
x = \sqrt{2.24 \times 10^{-11}}
\]
\[
x \approx 1.4967 \times 10^{-6}
\]
The pH is calculated using the concentration of \(\text{H}^+\):
\[
\text{pH} = -\log_{10}(x)
\]
\[
\text{pH} = -\log_{10}(1.4967 \times 10^{-6})
\]
\[
\text{pH} \approx 5.824
\]
The closest answer choice to our calculated pH is 5.75. Therefore, the answer is:
\(\boxed{5.75}\)
Answered
