The number of moles of NH3 , that must be added to 2 L of 0.80 M AgNO3

Question:

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 $\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}=1.0 \times 10^{8}$ ) is - (Nearest integer)

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

Solution:

Let moles added $=\mathrm{a}$

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

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

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

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

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