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Ionic Equilibrium question

2021 · 16 Mar · Shift 1 · Q15
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  5. /2021 · 16 Mar · Shift 1 · Q15

Ionic Equilibrium question

2021 · 16 Mar · Shift 1 · Q15

JEE MainChemistryIonic EquilibriumNumerical+4 / −1
Two salts A2XA_2XA2​X and MXMXMX have the same value of solubility product of 4.0 ×\times× 10 −-− 12. The ratio of their molar solubilities i.e. S(A2X)S(MX){{S({A_2}X)} \over {S(MX)}}S(MX)S(A2​X)​ = ‾\underline{\hspace{2cm}}​. (Round off to the Nearest Integer)
Numerical answer
View written solutionFree

Correct answer: 50

  1. Write the dissolution equilibria

For salt MXMXMX: MX(s)⇌M++X−MX(s) \rightleftharpoons M^+ + X^-MX(s)⇌M++X− If its molar solubility is s1s_1s1​, then [M+]=s1,[X−]=s1[M^+] = s_1, \quad [X^-] = s_1[M+]=s1​,[X−]=s1​ So, Ksp=[M+][X−]=s12K_{sp} = [M^+][X^-] = s_1^2Ksp​=[M+][X−]=s12​ Given: Ksp=4.0×10−12K_{sp} = 4.0 \times 10^{-12}Ksp​=4.0×10−12 Hence, s1=4.0×10−12=2.0×10−6s_1 = \sqrt{4.0 \times 10^{-12}} = 2.0 \times 10^{-6}s1​=4.0×10−12​=2.0×10−6

  1. For salt A2XA_2XA2​X

Its dissolution is: A2X(s)⇌2A++X2−A_2X(s) \rightleftharpoons 2A^+ + X^{2-}A2​X(s)⇌2A++X2− If its molar solubility is s2s_2s2​, then [A+]=2s2,[X2−]=s2[A^+] = 2s_2, \quad [X^{2-}] = s_2[A+]=2s2​,[X2−]=s2​ Thus, Ksp=[A+]2[X2−]=(2s2)2(s2)=4s23K_{sp} = [A^+]^2[X^{2-}] = (2s_2)^2(s_2) = 4s_2^3Ksp​=[A+]2[X2−]=(2s2​)2(s2​)=4s23​ Given: 4s23=4.0×10−124s_2^3 = 4.0 \times 10^{-12}4s23​=4.0×10−12 So, s23=10−12s_2^3 = 10^{-12}s23​=10−12 s2=10−4s_2 = 10^{-4}s2​=10−4

  1. Find the ratio

S(A2X)S(MX)=s2s1=10−42×10−6=50\frac{S(A_2X)}{S(MX)} = \frac{s_2}{s_1} = \frac{10^{-4}}{2 \times 10^{-6}} = 50S(MX)S(A2​X)​=s1​s2​​=2×10−610−4​=50

  1. Final answer

50\boxed{50}50​

  1. Comparison with stored correct answer

Stored correct answer = 505050

This matches the derived answer.

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