The number of moles of $\ce{NH3}$, that must be added to 2 L of $0.80~\mathrm{M}~\ce{AgNO3}$ in order to reduce the concentration of $\ce{Ag^+}$ ions to $5.0 \times 10^{-8}~\mathrm{M}$ ($K_{\text{formation}}$ for $[\ce{Ag(NH3)2}]^{+}=1.0 \times 10^{8}$) is ______. (Nearest integer) [Assume no volume change on adding $\ce{NH3}$]

JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Coordination Compounds

The number of moles of

NH X_{3}

, that must be added to 2 L of

0.80 M AgNO X_{3}

in order to reduce the concentration of

Ag X^{+}

ions to

5.0 \times 1 0^{- 8} M

(

K_{formation}

for

[Ag (NH X_{3}) X_{2}]^{+} = 1.0 \times 1 0^{8}

) is ______. (Nearest integer)
[Assume no volume change on adding

NH X_{3}

]

Answer: 4

The silver present is

0.80 \times 2 = 1.6

mol, and lowering

[A g^{+}]

5.0 \times 1 0^{- 8} M

converts essentially all of it to

[Ag (N H_{3})_{2}]^{+}

, whose concentration is therefore

1.6/2 = 0.8 M

. From

K_{f} = \frac{[[ Ag ( N H _{3} ) _{2} ] ^{+} ]}{[ A g ^{+} ] [ N H _{3} ] ^{2}}

[N H_{3}]^{2} = \frac{0.8}{( 1.0 \times 1 0 ^{8} ) ( 5.0 \times 1 0 ^{- 8} )} = 0.16

, so

[N H_{3}] = 0.4 M

and the free ammonia is

0.8

mol. Adding the

2 \times 1.6 = 3.2

mol tied up in the complex, the ammonia that must be added is

3.2 + 0.8 = 4

mol.

JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Integer answer