$\mathrm{pH}=\mathrm{pKa}+\log \frac{[\mathrm{Salt}]}{\mathrm{Acid}}$
$\mathrm{pH}=4.5+\log \frac{(\mathrm{Salt}]}{\mathrm{Acid}}$
As HA is $50 \%$ ionized so $[$ Salt $]=[$ Acid $]$
$\mathrm{pH}=4.5+\log (1)=4.5$
Now, $\mathrm{pH}+\mathrm{pOH}=14$
$\therefore \mathrm{pOH}=14-4.5=9.5$
$\left[\right.$ આપેલ : $pK _{ b }\left( NH _3\right)=4.745$
$\log 2=0.301$
$\log 3=0.477$
$T =298\,K ]$
[એસિટિક એસિડનો $\mathrm{pK}_{\mathrm{a}}$ $=4.75,$, એસિટિક એસિડનું આણ્વિય દળ$=60 \mathrm{g} / \mathrm{mol}, \log 3=0.4771]$
કદમાં થતો ફેરફાર અવગણો
(આપેલ $CaF _{2}$ નું મોલર દળ = $78 \,g mol ^{-1}$ )