The number of moles of NH_ 3 , that must be added to 2 ,~L of 0.8… - Vidyadip

MCQ

The number of moles of $\mathrm{NH}_{3}$, that must be added to $2 \,\mathrm{~L}$ of $0.80\, \mathrm{M}\, \mathrm{AgNO}_{3}$ in order to reduce the concentration of $\mathrm{Ag}^{+}$ions to $5.0 \times 10^{-8}\, \mathrm{M}\left(\mathrm{K}_{\text {formation }}\right.$ for $\left.\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}=1.0 \times 10^{8}\right)$ is ...... . (Nearest integer)

[Assume no volume change on adding $\mathrm{NH}_{3}$ ]

A
$16$
B
$5$
✓
$4$
D
$2$

✓

Answer

Correct option: C.

$4$

c
Let moles added $=\mathrm{a}$

$\mathrm{Ag}_{(\mathrm{ag} .)}^{+}+2 \mathrm{NH}_{3(\mathrm{ag} .)} \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2(\mathrm{aq} .)}^{+}$

$t=0 \quad \quad 0.8 \quad\quad\quad \left(\frac{a}{2}\right)$

$t=\infty \quad\quad 5 \times 10^{-8} \quad\left(\frac{a}{2}-1.6\right) \quad 0.8$

$\frac{0.8}{\left(5 \times 10^{-8}\right)\left(\frac{a}{2}-1.6\right)^{2}}=10^{8}$

$\frac{a}{2}-1.6=0.4 \Rightarrow a=4$

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