Question

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

Short answer

The Q_spvalue is greater than K_spfor ${\mathbf{\text{Cu(OH)}}}_{\mathbf{\text{2}}}$which is role="math" localid="1663569579426" $2.2\times {10}^{-20}$and because of that solid ${\mathbf{\text{Cu(OH)}}}_{\mathbf{\text{2}}}$will be formed.

Step-by-step solution

Relation between solubility product and ionic product

Ionic product expression; Q_spvalue is obtained when the concentrations of ions formed by dissolution of some compound are multiplied. When the solution is saturated Q_Spvalue is called K_spvalue (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_spequations.

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

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

Calculation

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

role="math" localid="1663570661037" $$\begin{gathered} \left[{\mathrm{OH}}^{-}\right]=\frac{\mathrm{m}\left(\mathrm{KOH}\right)}{\mathrm{M}\left(\mathrm{KOH}\right)} \\ =\frac{0.075\mathrm{g}}{56.1\mathrm{g}/\mathrm{mol}} \\ =1.34\times {10}^{-3}\mathrm{M} \end{gathered}$$

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_spand solving –

Square${\text{[OH}}^{\text{-}}\text{]}$as there aremoles ofin the equation.

role="math" localid="1663573637861" $$\begin{gathered} {\mathrm{Q}}_{\mathrm{sp}}=\left[{\mathrm{Cu}}^{2+}\right]\left[{\mathrm{OH}}^{-}\right] \\ =(0.001)·(1.34\times {10}^{-3}{)}^2 \\ =1.8\times {10}^{-9} \end{gathered}$$

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.