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Question: Three sparingly soluble salts have same solubility products are given below: \({\rm I}.{A_2}X\) \(...

Three sparingly soluble salts have same solubility products are given below:
I.A2X{\rm I}.{A_2}X II.AX{\rm I}{\rm I}.AX III.AX3{\rm I}{\rm I}{\rm I}.A{X_3}
Their solubility in a standard solution will be such that:
A: III>II>I{\rm I}{\rm I}{\rm I} > {\rm I}{\rm I} > {\rm I}
B: III>I>II{\rm I}{\rm I}{\rm I} > {\rm I} > {\rm I}{\rm I}
C: II>III>I{\rm I}{\rm I} > {\rm I}{\rm I}{\rm I} > {\rm I}
D: II>I>III{\rm I}{\rm I} > {\rm I} > {\rm I}{\rm I}{\rm I}

Explanation

Solution

Solubility product of an electrolyte at a particular temperature is defined as the product of the molar concentration of its ions in a saturated solution, each concentration raised to the power equal to the number of ions produced on dissociation. It is represented byKsp{K_{sp}}.
Formula used: AxByxA++yB{A_x}{B_y}\overset {} \leftrightarrows x{A^ + } + y{B^ - }
Ksp=[Ay+]x[Bx]y{K_{sp}} = {\left[ {{A^{y + }}} \right]^x}{\left[ {{B^{x - }}} \right]^y}
Where Ksp{K_{sp}}is solubility product,[Ay+]\left[ {{A^{y + }}} \right]is concentration ofAy+{A^{y + }} and[Bx]\left[ {{B^{x - }}} \right]is concentration ofBx{B^{x - }}.

Complete answer:
Sparingly soluble salts are those salts whose solubility is very low. The solubility of a substance is defined as the amount of substance which is soluble in 100mL100mL of water. The solubility product of an electrolyte at a particular temperature is defined as the product of the molar concentration of its ions in a saturated solution. Given salts have the same solubility product. Let their solubility productKsp{K_{sp}} be mm.
For salt A2X{A_2}X :
Solubility product Ksp=m{K_{sp}} = m (stated above)
Let solubility of salt be s1{s_1}
At equilibrium A2X2A++X{A_2}X\overset {} \leftrightarrows 2{A^ + } + {X^ - }
[A+]=[X2]=s1\left[ {{A^ + }} \right] = \left[ {{X^{2 - }}} \right] = {s_1} , x=2x = 2 and y=1y = 1 (according to the formula stated above)
Applying the above formula Ksp=[Ay+]x[Bx]y{K_{sp}} = {\left[ {{A^{y + }}} \right]^x}{\left[ {{B^{x - }}} \right]^y} (stated above)
m=[A+]2[X2]m = {\left[ {{A^ + }} \right]^2}\left[ {{X^{2 - }}} \right]
m=s12×s1m = s_1^2 \times {s_1}
m=s13m = s_1^3
s1=(m)13{s_1} = {\left( m \right)^{\dfrac{1}{3}}}
For saltAXAX :
Solubility product Ksp=m{K_{sp}} = m (stated above)
Let solubility of salt be s2{s_2}
At equilibrium AXA++XAX\overset {} \leftrightarrows {A^ + } + {X^ - }
[A+]=[X]=s2\left[ {{A^ + }} \right] = \left[ {{X^ - }} \right] = {s_2} , x=1x = 1 and y=1y = 1 (according to the formula stated above)
Applying the above formula Ksp=[Ay+]x[Bx]y{K_{sp}} = {\left[ {{A^{y + }}} \right]^x}{\left[ {{B^{x - }}} \right]^y} (stated above)
m=[A+][X]m = \left[ {{A^ + }} \right]\left[ {{X^ - }} \right]
m=s2×s2m = {s_2} \times {s_2}
m=s22m = s_2^2
s2=(m)12{s_2} = {\left( m \right)^{\dfrac{1}{2}}}
For salt AX3A{X_3} :
Solubility product Ksp=m{K_{sp}} = m (stated above)
Let solubility of salt bes3{s_3}
At equilibrium AX3A++3XA{X_3}\overset {} \leftrightarrows {A^ + } + 3{X^ - }
[A3+]=[X]=s3\left[ {{A^{3 + }}} \right] = \left[ {{X^ - }} \right] = {s_3} , x=1x = 1 and y=3y = 3 (according to the formula stated above)
Applying the above formula Ksp=[Ay+]x[Bx]y{K_{sp}} = {\left[ {{A^{y + }}} \right]^x}{\left[ {{B^{x - }}} \right]^y} (stated above)
m=[A3+][X]3m = \left[ {{A^{3 + }}} \right]{\left[ {{X^ - }} \right]^3}
m=s3×s33m = {s_3} \times s_3^3
m=s34m = s_3^4
s3=(m)14{s_3} = {\left( m \right)^{\dfrac{1}{4}}}
Nows1=(m)13{s_1} = {\left( m \right)^{\dfrac{1}{3}}},s2=(m)12{s_2} = {\left( m \right)^{\dfrac{1}{2}}},s3=(m)14{s_3} = {\left( m \right)^{\dfrac{1}{4}}}
From these values we can clearly see thats2>s1>s3{s_2} > {s_1} > {s_3} .s1{s_1}Is solubility ofA2X{A_2}X,s1{s_1}is solubility of AXAXand s3{s_3}is solubility of AX3A{X_3}. In the question A2X{A_2}X is denoted as I{\rm I}, AXAXis denoted as II{\rm I}{\rm I}and AX3A{X_3} is stated as III{\rm I}{\rm I}{\rm I}. This means II>I>III{\rm I}{\rm I} > {\rm I} > {\rm I}{\rm I}{\rm I} (as s2>s1>s3{s_2} > {s_1} > {s_3}and s1{s_1}Is solubility ofA2X{A_2}X,s1{s_1}is solubility of AXAXand s3{s_3}is solubility of AX3A{X_3}).

So, correct answer is option D that isII>I>III{\rm I}{\rm I} > {\rm I} > {\rm I}{\rm I}{\rm I}.

Note:
If the solution of a weak electrolyte, a strong electrolyte having a common ion is added, the ionization of the weak electrolyte is further suppressed. For example, addition of CH3COONaC{H_3}COONa to CH3COOHC{H_3}COOH solution suppresses the ionization of CH3COOHC{H_3}COOH.