Ammonium sulphate ${\left(N{H}_4\right)}_{2}S{O}_4$ and NaOH (1 mol of each) is mixed and volume of solution is 1 lit. The pH of this solution at  ${25}^{0}C$ is ‘x’. Then (x + 1.74) is $(pKbofN{H}_{4}OH=4.74)$

${\left(N{H}_4\right)}_{2}S{O}_4$  on reaction with NaOH can form a buffer form a buffer solution.
${\left(N{H}_4\right)}_{2}S{O}_4+2NaOH\to N{a}_{2}S{O}_4+H_{2}O+2N{H}_3$ 
1 mol                        1                        0           0
0.5                            0                                                  1
$pOH=pKb+\log \left[\frac{N{H}_4^{+}}{N{H}_3}\right] =4.74+\log \left[\frac{0.5\times 2}{1}\right]=4.74$