JEE MAIN - Chemistry (2021 - 20th July Evening Shift - No. 4)
A solution is 0.1 M in Cl$$-$$ and 0.001 M in CrO$$_4^{2 - }$$. Solid AgNO3 is gradually added to it. Assuming that the addition does not change in volume and Ksp(AgCl) = 1.7 $$\times$$ 10$$-$$10 M2 and Ksp(Ag2CrO4) = 1.9 $$\times$$ 10$$-$$12 M3.
Select correct statement from the following :
Select correct statement from the following :
AgCl precipitates first because its Ksp is high.
Ag2CrO4 precipitates first as its Ksp is low.
Ag2CrO4 precipitates first because the amount of Ag+ needed is low.
AgCl will precipitate first as the amount of Ag+ needed to precipitate is low.
Explanation
Conc. of Cl$$-$$ = 0.1 M = 10$$-$$1 M
Conc. of CrO$$_4^{2 - }$$ = 0.001 M = 10$$-$$3 M
Ksp(AgCl) = [Ag+][Cl$$-$$]
[Ag+]AgCl = $${{1.7 \times {{10}^{ - 10}}} \over {{{10}^{ - 1}}}} = 1.7 \times {10^{ - 9}}$$
$${K_{sp}}(A{g_2}Cr{O_4}) = {[A{g^ + }]^2}[CrO_4^{2 - }]$$
$$[A{g^ + }] = \sqrt {{{1.9 \times {{10}^{ - 12}}} \over {{{10}^{ - 3}}}}} = \sqrt {19} \times {10^{ - 4}}$$
$$\therefore$$ AgCl will be precipitate first
Conc. of CrO$$_4^{2 - }$$ = 0.001 M = 10$$-$$3 M
Ksp(AgCl) = [Ag+][Cl$$-$$]
[Ag+]AgCl = $${{1.7 \times {{10}^{ - 10}}} \over {{{10}^{ - 1}}}} = 1.7 \times {10^{ - 9}}$$
$${K_{sp}}(A{g_2}Cr{O_4}) = {[A{g^ + }]^2}[CrO_4^{2 - }]$$
$$[A{g^ + }] = \sqrt {{{1.9 \times {{10}^{ - 12}}} \over {{{10}^{ - 3}}}}} = \sqrt {19} \times {10^{ - 4}}$$
$$\therefore$$ AgCl will be precipitate first
Comments (0)
