Answer:
We know, \[pOH\]
of weak monoacidic base may be calculated as:
\[pOH=-\frac{1}{2}\log (C{{K}_{b}})\]
\[14-9.95=-\frac{1}{2}\log (0.005\times
{{k}_{b}})\]
\[-8.10=\log (0.005\times {{K}_{b}})\]
\[7.943\times {{10}^{-9}}=0.005\times {{K}_{b}}\]
\[{{K}_{b}}=1.588\times
{{10}^{-6}}\approx 1.6\times {{10}^{-6}}\]
\[p{{K}_{b}}=-{{\log }_{10}}{{K}_{b}}\]
\[=-\log 1.588\times {{10}^{-6}}\]
\[=5.799\]
\[\rho {{K}_{b}}=5.8\]
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