The ratio of volumes of $C{H}_{3}COOH0.1\left(N\right)$  to $C{H}_{3}COONa0.1\left(N\right)$  required to prepare a buffer solution of $pH$ 5. 74 is (Given, $p{K}_{a}ofC{H}_{3}COOH$  is 4.74)

Solution: The volumes of acetic acid and sodium acetate be $VL$ and V: respectively.
Given that normality of $C{H}_{3}COOH$  [Acetic acid] = 0.1 N 
Normality of   $C{H}_{3}COONa=0.1N$
                                                  $pH=5.74$
According to the Henderson’s equation – 
 $pH=p{K}_a+\log \frac{\left[C{H}_{3}COONa\right]}{\left[C{H}_{3}COOH\right]}$
 $5.74=4.74+\log \frac{0.1\times V\text{'}}{0.1\times V}$
 $5.74-4.74=\log \frac{V\text{'}}{V}$
 $1=\log \frac{V\text{'}}{V}$