A buffer solution has equal volumes of 0.2M NH4OH and 0.02M NH4CI . The pKb of the base is 5. The pH is

pOH=pKb+log⁡[ salt ][ base ]=5+log⁡0.020.2=5+log⁡110=5+(−1)=4pH=14−pOH=14−4=10.

A buffer solution has equal volumes of 0.2M NH4OH and 0.02M NH4CI . The pKb of the base is 5. The pH is

pOH=pKb+log⁡[ salt ][ base ]=5+log⁡0.020.2=5+log⁡110=5+(−1)=4pH=14−pOH=14−4=10.

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A buffer solution has equal volumes of 0.2M NH₄OH and 0.02M NH₄CI . The pK_b of the base is 5. The pH is

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$\begin{array}{l} pOH = {pK}_{b} + \log \frac{[salt]}{[base]} \\ = 5 + \log \frac{0 .02}{0 .2} = 5 + \log \frac{1}{10} = 5 + (- 1) = 4 \\ pH = 14 - pOH = 14 - 4 = 10 . \end{array}$

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A buffer solution has equal volumes of 0.2M NH4OH and 0.02M NH4CI . The pKb of the base is 5. The pH is

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A buffer solution has equal volumes of 0.2M NH₄OH and 0.02M NH₄CI . The pK_b of the base is 5. The pH is