What is the value of pOH if a buffer solution | vedclass

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(B) For a basic buffer,the Henderson-Hasselbalch equation is given by: $pOH = pK_b + \log \frac{[Salt]}{[Base]}$ Since equal volumes are mixed,the new

Vedclass · Accepted Answer

$4.83$

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(B) For a basic buffer,the Henderson-Hasselbalch equation is given by: $pOH = pK_b + \log \frac{[Salt]}{[Base]}$ Since equal volumes are mixed,the new concentrations are halved,but the ratio of $[Salt]/[Base]$ remains the same as the ratio of their initial concentrations. $[Salt] = [NH_4Cl] = 0.5 \ M$ $[Base] = [NH_4OH] = 0.4 \ M$ $pOH = 4.730 + \log \left( \frac{0.5}{0.4} \right)$ $pOH = 4.730 + \log(1.25)$ $pOH = 4.730 + 0.0969 \approx 4.8269 \approx 4.83$

What is the value of $pOH$ if a buffer solution is prepared by mixing equal volumes of $0.4 \ M$ $NH_4OH$ and $0.5 \ M$ $NH_4Cl$ solutions? $(pK_b = 4.730)$

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