Answer:
We know,
\[{{K}_{a}}\times
{{K}_{b}}={{K}_{w}}\]
\[{{K}_{a}}\frac{{{K}_{w}}}{{{K}_{b}}}=\frac{{{10}^{-14}}}{1.77\times
{{10}^{-5}}}=5.65\times {{10}^{-10}}\]
For the mono
acidic base:
\[\alpha
=\sqrt{\frac{{{K}_{b}}}{C}}=\sqrt{\frac{1.77\times {{10}^{-5}}}{0.05}}\]
= 0.0188
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