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Question

Chemistry Question on Equilibrium

Solubility product constant ( KspK_{sp} )of salts of types MX1,MX2andM3XMX_1, MX_ 2 \, and \, M_3 \, X at temperature -T are 4.0×108,3.2×10144.0 \times 10^ { - 8 } , \, 3.2 \times 10^{ - 14 } and 2.7×1015 2.7 \times 10^{ - 15 }, respectively. Solubilities (mol dm3^{- 3} ) of the salts at temperature T" are in the order

A

MX>MX2>M3XMX > MX_2 > M_3 X

B

M3X>MX2>MXM_3X > MX_2 > MX

C

MX2>M2X>MX2MX_2 > M_2 X > MX_2

D

MX>M3X>MX2MX > M_3 X > MX_2

Answer

MX>M3X>MX2MX > M_3 X > MX_2

Explanation

Solution

MX : Ksp=S2=4×108K_{ s p } = S^2 = 4 \times 10^{ - 8 }
S=2×104\Rightarrow S = 2 \times 10^{ - 4 }
MX2:Ksp=4S3=3.2×1014S=2×105MX_2 : K_{ s p } = 4 \, S^3 = 3.2 \times 10^{ - 14 } \Rightarrow S = 2 \times 10^{ - 5 }
M3X:Ksp=27S4=2.7×1015S=104M_3 \, X : K_{ sp } = 27 S^4 = 2.7 \times 10^{ - 15 } \Rightarrow S = 10^{ - 4 }
Order of solubility is MX>M3X>MX2MX > M_3 X > MX_2