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Calculate the pH of a solution made by mixing 0.1 M NH3 and 0.1 M (NH4)2SO4. (pKb of NH3 = 4.76)

Option: 1

7.2


Option: 2

6.84


Option: 3

8.94


Option: 4

4.6


Answers (1)

best_answer

\text { It is a basic buffer. }
\begin{array}{l}{\text { [ Base] = }\left[\mathrm{NH}_{3}\right]=0.1 \mathrm{M}} \\ {\begin{aligned}{}\text{\:[Salt }]=&\left[\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\right]=0.1 \times 2=0.2 \mathrm{M} \\ &\left[\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} \longrightarrow 2 \mathrm{NH}_{4}+\mathrm{SO}_{4}^{2-}\right] \\ \mathrm{pOH}=& \mathrm{p} K_{b}+\log \frac{[\mathrm{Salt}]}{[\mathrm{Base}]} \\=& 4.76+\log \frac{0.2 \mathrm{M}}{0.1 \mathrm{M}} \\=& 4.76+\log 2 \\=& 4.76+0.3=5.06 \\ \mathrm{pH}=& 14-5.03=8.94 \end{aligned}}\end{array}

Therefore, Option(3) is correct

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