20ml of 0.1 M H_ 2 SO _ 4 solution is added to 30mL of 0 .2 M NH _...

20ml of

{\text{0.1 M H}_{2}{SO}_{4}

solution is added to

30mL of 0 .2 M{NH}_{4}OH

solution. The

pH

of the resultant mixture is: \left[{pk}_{b}of{NH}_{4}OH = 4 .7\right] 20ml of 0.1 M H_ 2 SO _ 4 solution is added to 30mL of 0 .2 M NH _ 4 OH solution. The pH of the resultant mixture is: [

20ml of 0.1 M H_ 2 SO _ 4 solution is added to 30mL of 0 .2 M NH _ 4 OH solution. The pH of the resultant mixture is: [

Options

A $5.0$
B $5.2$
C $9.4$
D $9.0$

Correct answer

D. $9.0$

Step-by-step solution

Millimoles of ${H}_{2}S{O}_{4} = 20 \times 0.1 = 2$

Millimoles of $N{H}_{4}OH = 30 \times 0.2 = 6$

20ml of 0.1 M H_ 2 SO _ 4 solution is added to 30mL of 0 .2 M NH _ 4 OH solution. The pH of the resultant mixture is: [

Hence, the solution will act as a basic buffer.

$pOH = p{K}_{b} + log\dfrac{\left[salt\right]}{\left[Base\right]}$

\left[N{H}_{4}^{+}\right] = 2 \times \left[{\left(N{H}_{4}\right)}_{2}S{O}_{4}\right]; {\left[N{H}_{4}\right]}^{+} = 4 $ $ pOH = p{K}_{b} + log\dfrac{\left[N{H}_{4}^{+}\right]}{\left[N{H}_{3}\right]} $ $ = 4.7 + log\dfrac{4}{2} $ $ = 4.7 + 0.3010 = 5.010 $ $ pH = 14 - pOH = 9

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