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In qualitative analysis, the metals of group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains Ag+ and Pb2+ at a concentration is 0.10 M. Aqueous HCl is added to this solution until the Clconcentration is 0.10 M. What will the concentrations of Ag+ and Pb2+ be at equilibrium?

\mathrm{\left( K _{ sp } \text { for } AgCl =1.8 \times 10^{-10}, K _{ sp } \text { for } PbCl _2=1.7 \times 10^{-5}\right)}

Option: 1

[Ag+] = 1.8 x 10-9 M

[Pb2+] = 1.7 x 10-3 M


Option: 2

[Ag+] = 1.8 x 10-11 M

[Pb2+] = 8.5 x 10-4 M


Option: 3

[Ag+] = 1.8 x 10-7 M

[Pb2+] = 1.7 x 10-6 M


Option: 4

[Ag+] = 1.8 x 10-11 M

[Pb2+] = 8.5 x 10-5 M


Answers (1)

best_answer

Ksp of AgCl and PbCl2 is given.

So,

\\\mathrm{K_{s p}}[\mathrm{AgCl}]=\left[\mathrm{Ag}^{+}\right][\mathrm{Cl^-}]\\\\ 1.8 \times 10^{-10} = \left[\mathrm{Ag}^{+}\right]\times (10^{-1}) \\ \\ \left[\mathrm{Ag}^{+}\right]=\frac{1.8 \times 10^{-10}}{10^{-1}}=1.8 \times 10^{-9} \mathrm{M}

 

\\\mathrm{K_{s p}}[\left[\mathrm{PbCl}_{2}\right]=\left[\mathrm{Pb}^{2+}\right][\mathrm{Cl^-}]^{2}\\\\ 1.7 \times 10^{-5} = \left[\mathrm{Pb}^{2+}\right] \times (10^{-1} )^2 \\ \\ \left[\mathrm{Pb}^{2+}\right]=\frac{1.7 \times 10^{-5}}{10^{-1} \times 10^{-1}}=1.7 \times 10^{-3} \mathrm{M}

Therefore, 1st option is correct.

Posted by

jitender.kumar

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