Chapter 17, Problem 17.118SP

Interpretation Introduction

Interpretation:

The concentrations of ions in solution obtained when $100mLof0.1MCuS{O}_4$ and $100mLof0.1MBa{\left(OH\right)}_2$ are mixed should be calculated.

Concept Introduction:

Solubility is defined as the maximum quantity of solute dissolved in a given amount of solvent to make a saturated solution at a particular temperature.

Molar solubility is the number of moles of a solute that can be dissolved in one liter of a solution. It is expressed as mol/L or M (molarity).

Consider a general reaction:

  ${\text{M}}_{\text{n}}{\text{X}}_{\text{m}}\text{(s)}\stackrel{}{\rightleftharpoons }\text{n}{\text{M}}^{\text{m+}}\text{(aq)+m}{\text{X}}^{\text{n+}}\text{(aq)}$

The relation between solubility product and molar solubility is as follows:

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

Here

The solubility product of salt is ${\text{K}}_{\text{sp}}$.

The molar solubility of ${\text{M}}^{\text{m+}}$ ion is $\left[{\text{M}}^{\text{m+}}\right]$.

The molar solubility of ${\text{X}}^{\text{n-}}$ ion is $\left[{\text{X}}^{\text{n-}}\right]$.

Molarity: The concentration for solutions is expressed in terms of molarity as follows,

  $\text{Molarity = }\frac{\text{No}\text{. of moles of solute}}{\text{Volume of solution in L}}$

Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in $\text{12}\text{g}$ of ${}^{\text{12}}\text{C}$.

From given mass of substance moles could be calculated by using the following formula,

  $\text{Moles}\text{of}\text{substance }\text{= }\frac{\text{Given}\text{mass}\text{of}\text{substance}}{\text{Molecular}\text{mass}}$

Answer to Problem 17.118SP

The concentrations of ions in solution are as follows,

    $\begin{array}{c}\left[B{a}^{2+}\right]=1.0\times {10}^{-5}M\\ {\left[S{O}_4^2{}^{-}\right]}^{}=1.0\times {10}^{-5}M\end{array}$

Explanation of Solution

When $100mLof0.1MCuS{O}_4$ and $100mLof0.1MBa{\left(OH\right)}_2$ are mixed, $Cu{\left(OH\right)}_{2}andBaS{O}_4$ are precipitated out of the solution.

  $\begin{array}{c}CuS{O}_4\\ \text{No}\text{. of moles of solute}=\text{Molarity}\times \text{Volume of solution in L}\\ \text{=}\text{0}\text{.1}\times \text{100}\times {\text{10}}^{\text{-3}}L\\ =0.01mol\end{array}$

    $\begin{array}{l}CuS{O}_4\left(aq\right)+Ba{\left(OH\right)}_2\left(aq\right)\to Cu{\left(OH\right)}_2\left(s\right)+BaS{O}_4\left(s\right)\\ Initial\left(mol\right):0.010.0100\\ Change\left(mol\right):\text{-0}\text{.01}\text{-0}\text{.01}+\text{0}\text{.01}+\text{0}\text{.01}\\ ----------------------------------------\\ Final\left(mol\right):000.010.01\end{array}$

Now using the each precipitate dissociation in water the concentration of the ions are calculated as follows,

    $\begin{array}{l}Cu{\left(OH\right)}_2\left(s\right)\rightleftharpoons C{u}^{2+}\left(aq\right)+2O{H}^{-}\left(aq\right)\\ Initial\left(mol\right):s00\\ Change\left(mol\right):\text{-s}+\text{s}+2\text{s}\\ ----------------------------------------\\ Equilibrium\left(mol\right):s2\text{s}\end{array}$

    $\begin{array}{c}{K}_{sp}=\left[C{u}^{2+}\right]{\left[O{H}^{-}\right]}^2\\ 2.2\times {10}^{-20}=s.{\left(2s\right)}^2\\ 2.2\times {10}^{-20}=4s^3\\ s=1.8\times {10}^{-7}M\end{array}$

    $\begin{array}{c}\left[C{u}^{2+}\right]=s=1.8\times {10}^{-7}M\\ {\left[O{H}^{-}\right]}^{}=2s=3.6\times {10}^{-7}M\end{array}$

    $\begin{array}{l}BaS{O}_4\left(s\right)\rightleftharpoons B{a}^{2+}\left(aq\right)+S{O}_4^{2-}\left(aq\right)\\ Initial\left(mol\right):s00\\ Change\left(mol\right):\text{-s}+\text{s}+\text{s}\\ ----------------------------------------\\ Equilibrium\left(mol\right):s\text{s}\end{array}$

    $\begin{array}{c}{K}_{sp}=\left[B{a}^{2+}\right]\left[S{O}_4^{2-}\right]\\ 1.1\times {10}^{-10}=s^2\\ s=\sqrt{1.1\times {10}^{-10}}\\ =1.0\times {10}^{-5}M\end{array}$

    $\begin{array}{c}\left[B{a}^{2+}\right]=s=1.0\times {10}^{-5}M\\ {\left[S{O}_4^2{}^{-}\right]}^{}=s=1.0\times {10}^{-5}M\end{array}$