The $\mathrm{pH}$ of solution can be given by

The $\mathrm{pH}$ of solution can be given by
  1. $\mathrm{pK}_{\mathrm{w}}+\log \mathrm{C}_{3}$
  2. $\mathrm{pK}_{\mathrm{w}}-\frac{1}{2} \mathrm{p} \mathrm{K}_{\mathrm{b}}+\frac{1}{2} \log \mathrm{C}_{2}$
  3. $\mathrm{pK}_{\mathrm{a}}+\frac{1}{2} \log \mathrm{C}_{2}+\frac{1}{2} \mathrm{p} \mathrm{K}_{\mathrm{b}}$
  4. All of these

Solution

$\left[\mathrm{OH}^{-}ight]=\mathrm{C}_{3}$ $\mathrm{C}_{2} \alpha=\mathrm{C}_{2} \sqrt{\frac{\mathrm{K}_{\mathrm{b}}}{\mathrm{C}_{2}}}=\sqrt{\mathrm{K}_{\mathrm{b}} \mathrm{C}_{2}}$
$\mathrm{pOH}=-\log \mathrm{C}_{3}=-\frac{1}{2}\left[\log \mathrm{K}_{\mathrm{b}}+\log \mathrm{C}_{2}ight]$
$=\frac{1}{2} \mathrm{pK}_{\mathrm{a}}-\frac{1}{2} \log \mathrm{C}_{2}$
$\mathrm{pH}+\mathrm{pOH}=\mathrm{pK}_{\mathrm{w}}$ and $\mathrm{pK}_{\mathrm{a}}+\mathrm{pK}_{\mathrm{b}}=\mathrm{pK}_{\mathrm{w}}$ ,

Asked in: JEE-TOPICTESTS-CHEMISTRY

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