Question

To prepare a buffer of pH 8.26 amount of $({NH}_4{)}_2{SO}_4$ to be added to 500 mL of 0.01 M ${NH}_{4}OH$ solution is : $\left[p{K}_a\right({NH}_4^{+})=9.26]$

• A. 0.05 mole
• B. 0.025 mole
• C. 0.10 mole
• D. 0.005 mole

Answer 1

Answer: (B)

0.025 mole

Let $(N{H}_4{)}_{2}S{O}_4$ to be added to 500 mL solution = x mol

$\left[N{H}_4^{+}\right]=4xM$

$pH=8.26$

$pOH=14-8.26=5.74$

$p{K}_b=14-9.26=4.74$

$pOH=p{K}_b+log\frac{\left[N{H}_4^{+}\right]}{\left[N{H}_{4}OH\right]}$

$5.74=4.74+log\frac{4x}{0.01}$

$log4x\times {10}^2=1$

$4x\times {10}^2=1$

$x=\frac{1}{40}mol=0.025mol$