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${\rm I}.{A_2}X$ ${\rm I}{\rm I}.AX$ ${\rm I}{\rm I}{\rm I}.A{X_3}$

Their solubility in a standard solution will be such that:

A: ${\rm I}{\rm I}{\rm I} > {\rm I}{\rm I} > {\rm I}$

B: ${\rm I}{\rm I}{\rm I} > {\rm I} > {\rm I}{\rm I}$

C: ${\rm I}{\rm I} > {\rm I}{\rm I}{\rm I} > {\rm I}$

D: ${\rm I}{\rm I} > {\rm I} > {\rm I}{\rm I}{\rm I}$

Answer

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Formula used: ${A_x}{B_y}\overset {} \leftrightarrows x{A^ + } + y{B^ - }$

${K_{sp}} = {\left[ {{A^{y + }}} \right]^x}{\left[ {{B^{x - }}} \right]^y}$

Where ${K_{sp}}$is solubility product,$\left[ {{A^{y + }}} \right]$is concentration of${A^{y + }}$ and$\left[ {{B^{x - }}} \right]$is concentration of${B^{x - }}$.

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 $100mL$ 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 product${K_{sp}}$ be $m$.

For salt ${A_2}X$ :

Solubility product ${K_{sp}} = m$ (stated above)

Let solubility of salt be ${s_1}$

At equilibrium ${A_2}X\overset {} \leftrightarrows 2{A^ + } + {X^ - }$

$\left[ {{A^ + }} \right] = \left[ {{X^{2 - }}} \right] = {s_1}$ , $x = 2$ and $y = 1$ (according to the formula stated above)

Applying the above formula ${K_{sp}} = {\left[ {{A^{y + }}} \right]^x}{\left[ {{B^{x - }}} \right]^y}$ (stated above)

$m = {\left[ {{A^ + }} \right]^2}\left[ {{X^{2 - }}} \right]$

$m = s_1^2 \times {s_1}$

$m = s_1^3$

${s_1} = {\left( m \right)^{\dfrac{1}{3}}}$

For salt$AX$ :

Solubility product ${K_{sp}} = m$ (stated above)

Let solubility of salt be ${s_2}$

At equilibrium $AX\overset {} \leftrightarrows {A^ + } + {X^ - }$

$\left[ {{A^ + }} \right] = \left[ {{X^ - }} \right] = {s_2}$ , $x = 1$ and $y = 1$ (according to the formula stated above)

Applying the above formula ${K_{sp}} = {\left[ {{A^{y + }}} \right]^x}{\left[ {{B^{x - }}} \right]^y}$ (stated above)

$m = \left[ {{A^ + }} \right]\left[ {{X^ - }} \right]$

$m = {s_2} \times {s_2}$

$m = s_2^2$

${s_2} = {\left( m \right)^{\dfrac{1}{2}}}$

For salt $A{X_3}$ :

Solubility product ${K_{sp}} = m$ (stated above)

Let solubility of salt be${s_3}$

At equilibrium $A{X_3}\overset {} \leftrightarrows {A^ + } + 3{X^ - }$

$\left[ {{A^{3 + }}} \right] = \left[ {{X^ - }} \right] = {s_3}$ , $x = 1$ and $y = 3$ (according to the formula stated above)

Applying the above formula ${K_{sp}} = {\left[ {{A^{y + }}} \right]^x}{\left[ {{B^{x - }}} \right]^y}$ (stated above)

$m = \left[ {{A^{3 + }}} \right]{\left[ {{X^ - }} \right]^3}$

$m = {s_3} \times s_3^3$

$m = s_3^4$

${s_3} = {\left( m \right)^{\dfrac{1}{4}}}$

Now${s_1} = {\left( m \right)^{\dfrac{1}{3}}}$,${s_2} = {\left( m \right)^{\dfrac{1}{2}}}$,${s_3} = {\left( m \right)^{\dfrac{1}{4}}}$

From these values we can clearly see that${s_2} > {s_1} > {s_3}$ .${s_1}$Is solubility of${A_2}X$,${s_1}$is solubility of $AX$and ${s_3}$is solubility of $A{X_3}$. In the question ${A_2}X$ is denoted as ${\rm I}$, $AX$is denoted as ${\rm I}{\rm I}$and $A{X_3}$ is stated as ${\rm I}{\rm I}{\rm I}$. This means ${\rm I}{\rm I} > {\rm I} > {\rm I}{\rm I}{\rm I}$ (as ${s_2} > {s_1} > {s_3}$and ${s_1}$Is solubility of${A_2}X$,${s_1}$is solubility of $AX$and ${s_3}$is solubility of $A{X_3}$).

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

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 $C{H_3}COONa$ to $C{H_3}COOH$ solution suppresses the ionization of $C{H_3}COOH$.