Q19.86P - Chemistry: Molecular Nature Of Matter And Change

Question

Does any solid ${\mathbf{\text{Cu(OH)}}}_{\mathbf{2}}$form when $\mathbf{\text{0.075 g}}$ of $\mathbf{\text{KOH}}$ is dissolved in  of $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{M}\mathbf{Cu}\mathbf{(}{\mathbf{NO}}_{\mathbf{3}}{\mathbf{)}}_{\mathbf{2}}$ ?

Step-by-Step Solution

Answer

The Q_sp value is greater than K_sp for ${\mathbf{\text{Cu(OH)}}}_{\mathbf{\text{2}}}$which is $2.2\times {10}^{-20}$and because of that solid ${\mathbf{\text{Cu(OH)}}}_{\mathbf{\text{2}}}$will be formed.

Relation between solubility product and ionic product

Ionic product expression; Q_sp value is obtained when the concentrations of ions formed by dissolution of some compound are multiplied. When the solution is saturated Q_Sp value is called K_sp value (solubility-product constant).

${\mathbf{\text{MX}}}_{\mathbf{\text{2}}}\mathbf{\to }{\mathbf{\text{M}}}^{\mathbf{\text{2 + }}}{\mathbf{\text{ + 2X}}}^{\mathbf{\text{ - }}}$

Solid and liquid state is not included in the K_sp equations.

${\mathbf{\text{K}}}_{\mathbf{\text{sp}}}{\mathbf{\text{ = [M}}}^{\mathbf{\text{2 + }}}{\mathbf{\text{][X}}}^{\mathbf{\text{ - }}}{\mathbf{\text{]}}}^{\mathbf{\text{2}}}$

S (Molar solubility) is equal to the concentration of one mol of the ion formed by dissolution of some compound.

If Q_sp >  K_sp, then precipitate will be formed.

Given Data

Given,

• Molar Mass of is: $\text{0.75 g}$
• Moles of  is: $1.0\times {10}^{-3}\mathrm{M}$
• Volume of is: $\text{1.0 L}$

Calculation

First, calculate the moles of ${\text{OH}}^{\text{ - }}$–

$$\begin{gathered} \left[{\mathrm{OH}}^{-}\right]=\frac{\mathrm{m}\left(\mathrm{KOH}\right)}{\mathrm{M}\left(\mathrm{KOH}\right)} \\ =\frac{0.075\mathrm{g}}{56.1\mathrm{g}/\mathrm{mol}} \\ =1.34\times {10}^{-3} \mathrm{M} \end{gathered}$$

Now write the dissolution equation for  ${\text{Cu(OH)}}_{\text{2}}$–

${\text{Cu(OH)}}_{\text{2}}\text{(s)}\rightleftharpoons {\text{Cu}}^{\text{2 + }}{\text{(aq) + 2OH}}^{\text{ - }}\text{(aq)}$

In this case, the initial concentration of ${\text{Cu}}^{\text{2 + }}$ion is, as there is so the equilibrium concentrations will be –

$$\begin{gathered} \left[{\mathrm{Cu}}^{2+}\right]=0.001\mathrm{M} \\ [{\mathrm{OH}}^{}-]=1.34\times {10}^{-3}\mathrm{M} \end{gathered}$$

Rearranging the equation of K_sp and solving –

Square ${\text{[OH}}^{\text{ - }}\text{]}$ as there are  moles of  in the equation.

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{sp}}=\left[{\mathrm{Cu}}^{2+}\right]\left[{\mathrm{OH}}^{-}\right] \\ =(0.001)·(1.34\times {10}^{-3}{)}^2 \\ =1.8\times {10}^{-9} \end{gathered}$$

The K_sp value for $\mathrm{Cu}{\left(\mathrm{OH}\right)}_2$ is –

${\mathrm{K}}_{\mathrm{sp}}=2.2\times {10}^{-20}$

 

Therefore, ionic product is greater than solubility product. Q_sp > K_sp. Hence, the precipitate of $\mathrm{Cu}{\left(\mathrm{OH}\right)}_2$ will be formed.