Chapter EQ, Problem 24PQ

Interpretation Introduction

Interpretation:

When $\text{100 mL}$ of $\text{0}\text{.06 M Mg}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is added to $50mL$ of $0.06MN{a}_2{C}_2{O}_4$ solution.

The result of the reaction has to be given.

Concept introduction:

The solubility product (K^sp): Solubility product is defined as the concentration of the dissolved ion’s rise to the power of its stoichiometric coefficients in an equilibrium condition.

Consider a general reaction:

    $M_n{X}_m(s)\rightleftharpoons n{M}^{m+}(aq)+m{X}^{n-}(aq)$

The solubility product is expressed as follows:

    $K_{sp}={\left[M^{m+}\right]}^n{\left[X^{n-}\right]}^m$

Here

Solubility product = K^sp,

Explanation of Solution

Given,

$\text{100 mL}$ (or) 0.1 L of $\text{0}\text{.06 M Mg}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is added to $50mL$ (or) 0.05 L of $0.06MN{a}_2{C}_2{O}_4$. $Ksp$ of $Mg{C}_2{O}_4(s)$ is $8.6\times {10}^{-5}$.

Reason for the correct option.

The dissociation of $Mg{C}_2{O}_4(s)$ is given below,

    ${\text{MgC}}_{\text{2}}{\text{O}}_{\text{4}}\text{(s)}\underset{}{\rightleftharpoons }{\text{Mg}}^{\text{2+}}\text{+}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{2-}}$

The Qsp value is calculated as follows,

        $\text{Qsp}\text{=}\left[{\text{Mg}}^{\text{2+}}\right]\left[{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{2-}}\right]$

The ions present in the solution is calculated as follows,

    $\begin{array}{l}\left[{\text{Mg}}^{\text{2+}}\right]\text{ions}\text{in}\text{solutions}\text{=}\frac{\text{0}\text{.06}\text{M×0}\text{.1}\text{L}}{\text{0}\text{.1}\text{L}\text{+}\text{0}\text{.050}\text{L}}\\ \left[{\text{Mg}}^{\text{2+}}\right]\text{ions}\text{in}\text{solutions}\text{=}\text{0}\text{.04}\text{M}\end{array}$

  $\begin{array}{l}\left[{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{2-}}\right]\text{ions}\text{in}\text{solutions}\text{=}\frac{\text{0}\text{.06}\text{M×0}\text{.050}\text{L}}{\text{0}\text{.1}\text{L}\text{+}\text{0}\text{.050}\text{L}}\\ \left[{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{2-}}\right]\text{ions}\text{in}\text{solutions}\text{=}\text{0}\text{.02}\text{M}\end{array}$

The Qsp value is given below,

  $\begin{array}{l}\text{Qsp}\text{=}\left[{\text{Mg}}^{\text{2+}}\right]\left[{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{2-}}\right]\\ \\ \text{Qsp}\text{=}\left[\text{0}\text{.04}\text{M}\right]\left[\text{0}\text{.02}\text{M}\right]\\ \\ \text{Qsp}\text{=}8.0{\text{×10}}^{\text{-4}}\text{M}\end{array}$

Qsp value is greater than the $Ksp$ of $Mg{C}_2{O}_4(s)$, therefore, precipitation will be form and the reactant will be there in the solutions.

When comparing the ${\text{Mg}}^{\text{2+}}$ ions and ${\text{C}}_{\text{2}}{\text{O}}_{\text{4}}^{\text{2-}}$ ions in solution, the number of ions of ${\text{Mg}}^{\text{2+}}$ is more in solution.

Hence, option (b) precipitation will form and ${\text{Mg}}^{\text{2+}}$ ions remain in solution is correct answer.

Reason for the incorrect option:

According to the calculation, the rest of the options (a), (c) and (d) are wrong answer.