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Chemistry Question on Law Of Chemical Equilibrium And Equilibrium Constant

The precipitate of CaF2, (Ksp=1.7×10−10)aF_2, \, ( K_{sp} = 1.7 \times 10^{ - 10 } ) is obtained, when equal volumes of which of the following are mixed?

A

10−4 M Ca2++10−4 M F−10^{ - 4 } \, M \, Ca^{ 2 + } + 10^{ - 4 } \, M \, F^-

B

10−2 M Ca2++10−3 M F−10^{ - 2 } \, M \, Ca^{ 2 + } + 10^{ - 3 } \, M \, F^-

C

10−5 M Ca2++10−3 M F−10^{ - 5 } \, M \, Ca^{ 2 + } + 10^{ - 3 } \, M \, F^-

D

10−3 M Ca2++10−5 M F−10^{ - 3 } \, M \, Ca^{ 2 + } + 10^{ - 5 } \, M \, F^-

Answer

10−2 M Ca2++10−3 M F−10^{ - 2 } \, M \, Ca^{ 2 + } + 10^{ - 3 } \, M \, F^-

Explanation

Solution

For precipitation reaction. QW>KspQ_W > K_{ sp }
QW=[Ca2+][F−]2=(10−22)×(10−32)Q_W = [ Ca^{ 2 +} ][ F^- ]^2 = \bigg( \frac{ 10^ { - 2 }}{ 2} \bigg) \times \bigg( \frac{ 10^ { - 3 }}{ 2} \bigg)
= 1.25×10−9>Ks1.25 \times 10^{ - 9 } > K_{ s} precipitate will be formed