The number of moles of NH_3, that must be added to 2 L of 0.80MAgNO_3 in order to reduce the concentration of Ag^(+) ions to 5.0×10^(-8)MK_(formation ) for AgNH3^(2+=1.0×108) is __________. (Nearest integer) [Assume no volume change on adding NH_3]

Correct answer: 4. Let no. moles of NH_3 added =x Ag^(+)(aq)+2NH_3(aq)⇌AgNH3^(2+(aq)t=0 0.8 x2t=∞ 5×10-8 x2-1.6 0.8Keq=AgNH32+Ag+NH32) x/2-1.6=0.4 x=4

Q29 / 90 · JEE Main 2021 (27 Aug Shift 1) · Chemistry

The number of moles of ${NH}_{3},$ that must be added to $2 L$ of $0.80 {MAgNO}_{3}$ in order to reduce the concentration of ${Ag}^{+}$ ions to $5.0 \times 10^{- 8} M (K_{formation}$ for ${[Ag {({NH}_{3})}_{2}]}^{+} = 1.0 \times 10^{8})$ is __________. (Nearest integer)
[Assume no volume change on adding ${NH}_{3}$ ]

Correct answer: 4

Solution

Let no. moles of ${NH}_{3}$ added $= x$

${Ag}^{+} (aq) + 2 {NH}_{3} (aq) ⇌ {[Ag {({NH}_{3})}_{2}]}^{+} (aq)$

$t = 0$ $0.8$ $\frac{x}{2}$

$t = \infty$ $5 \times 10^{- 8}$ $(\frac{x}{2} - 1.6)$ $0.8$

$K_{eq} = \frac{[{[Ag {({NH}_{3})}_{2}]}^{+}]}{[{Ag}^{+}] {[{NH}_{3}]}^{2}}$

$\frac{x}{2} - 1.6 = 0.4$

$x = 4$

Hence, the correct answer is 4.

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The number of moles of NH3, that must be added to 2 L of 0.80MAgNO3 in order to reduce the concentration of Ag+ ions to 5.0×10-8MKformation for AgNH32+=1.0×108 is __________. (Nearest integer)[Assume no volume change on adding NH3]

Solution

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