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Answer :
The pOH of the 0.045 M HZ solution is approximately 3.54.
Given the Ka value of HZ as [tex]\(1.55 \times 10^{-4}\),[/tex] we can determine the concentration of [tex]\(H^+\)[/tex] ions in the solution using the dissociation equilibrium of the weak acid:
[tex]\[K_a = \frac{[H^+][Z^-]}{[HZ]}\][/tex]
Since HZ is a weak acid, we can approximate that the concentration of[tex]\(Z^-\) ions (\([Z^-]\))[/tex] is equal to the concentration of [tex]\(H^+\)[/tex]ions [tex](\([H^+]\))[/tex] produced by the dissociation of HZ. Hence, we can rewrite the equation as:
[tex]\[K_a = \frac{[H^+]^2}{[HZ]}\][/tex]
Solving for [tex]\([H^+]\),[/tex] we find:
[tex]\[ [H^+] = \sqrt{K_a \times [HZ]} = \sqrt{1.55 \times 10^{-4} \times 0.045} \approx 1.57 \times 10^{-3} \text{ M}\][/tex]
The pOH can be calculated using the relationship between pH and [tex]\(H^+\) concentration (\([H^+]\)), which is \(pH = -\log([H^+])\). Since \(pH + pOH = 14\),[/tex] we can find the pOH:
[tex]\[pOH = 14 - pH = 14 - (-\log(1.57 \times 10^{-3})) \approx 3.54\][/tex]
Thus, the pOH of the solution is approximately 3.54.
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