Q22.83CP - Chemistry: Molecular Nature Of Matter And Change

Question

Even though most metal sulphides are sparingly soluble in water, their solubilities differ by several orders of magnitude. This difference is sometimes used to separate the metals by con pH. Use the following data to find the pH at which you can separate $\mathbf{\text{0.10}}$ M  ${\mathbf{\text{Cu}}}^{\mathbf{\text{2 + }}}$and $\mathbf{\text{0.10}}$ M${\mathbf{\text{Ni}}}^{\mathbf{\text{2 + }}}$ : Saturated ${\mathbf{\text{H}}}_{\mathbf{\text{2}}}\mathbf{\text{S}}$=0.10M

$$\begin{gathered} {\mathbf{\text{K}}}_{\mathbf{\text{a1}}}{\mathbf{\text{ of H}}}_{\mathbf{\text{2}}}{\mathbf{\text{S = 9×10}}}^{\mathbf{-}\mathbf{8}} \\ {\mathbf{\text{K}}}_{\mathbf{\text{a2}}}{\mathbf{\text{ of H}}}_{\mathbf{\text{2}}}\mathbf{\text{S = 1}}{\mathbf{\times 10}}^{\mathbf{-}\mathbf{17}} \\ {\mathbf{\text{K}}}_{\mathbf{\text{sp}}}\mathbf{\text{ of NiS=1.1}}{\mathbf{\times 10}}^{\mathbf{-}\mathbf{18}} \\ {\mathbf{\text{K}}}_{\mathbf{\text{sp}}}\mathbf{\text{ of CuS = 8}}{\mathbf{\times 10}}^{\mathbf{-}\mathbf{34}} \end{gathered}$$

Step-by-Step Solution

Answer

The pH value is obtained as:$\text{pH = 4.04}$ .

Step 1: Define Element

An element is a pure material made up entirely of atoms with the same number of protons in their nuclei, as defined by chemistry. Chemical elements, unlike chemical compounds, cannot be broken down into smaller substances by chemical reactions.

Step 2: Evaluate the value of pH

CuS will precipitate at a lower concentration of copper and sulphide ions because NiS is more soluble than CuS.

$$\begin{gathered} \text{CuS(s)}\to {\text{Cu}}^{\text{2 + }}{\text{(aq) + S}}^{\text{2 - }}\text{(aq)} \\ {\text{K}}_{\text{sp}}{\text{ = 8×10}}^{-34} \\ \text{NiS(s)}\to {\text{Ni}}^{\text{2 + }}{\text{(aq) + S}}^{\text{2 - }}\text{(aq)} \\ {\text{K}}_{\text{sp}}{\text{ = 1.1×10}}^{-18} \\ {\text{K}}_{\text{sp}}\text{(NiS) = }\left[{\text{Ni}}^{\text{2 + }}\right]\left[{\text{S}}^{\text{2 - }}\right] \\ \left[{\text{S}}^{\text{2 - }}\right]\text{ = }\frac{{\text{K}}_{\text{sp}}\text{(NiS)}}{\left[{\text{Ni}}^{\text{2 + }}\right]} \\ \left[{\text{S}}^{\text{2 - }}\right]\text{ = }\frac{{\mathrm{1.1\times 10}}^{-18}}{0.10} \\ \left[{\text{S}}^{\text{2 - }}\right]\text{ = }{\mathrm{1.1\times 10}}^{-17} M \end{gathered}$$

To prevent NiS from precipitating, the concentration of$S^{2-}$ ions must be kept below ${\mathrm{1.1\times 10}}^{-17} M$so that only CuS precipitates.

The next step is to determine the pH at which the sulphide ion concentration is less than ${\mathrm{1.1\times 10}}^{-17}$ M. Consider the following dissociation constants for ${\text{H}}_{\text{2}}\text{S}$:

$$\begin{gathered} {\text{H}}_{\text{2}}\text{S(s)}\leftrightarrow {\text{HS}}^{\text{ - }}{\text{(aq) + H}}^{\text{ + }}\text{(aq)} \\ {\text{K}}_{\text{a1}}{\text{ = 9×10}}^{-8} \\ {\text{HS}}^{\text{ - }}\text{(aq)}\leftrightarrow {\text{S}}^{\text{2 - }}{\text{(aq) + H}}^{\text{ + }}\text{(aq)} \\ {\text{K}}_{\text{a2}}{\text{ =1.1×10}}^{-17} \\ {\text{H}}_{\text{2}}\text{S(aq)}\leftrightarrow {\text{S}}^{\text{2 - }}{\text{(aq) + 2H}}^{\text{ + }}\text{(aq)} \\ {\text{K}}_{\text{a}}\text{ = }\frac{\left[{\text{S}}^{\text{2 - }}\right]{\left[{\text{H}}^{\text{ + }}\right]}^{\text{2}}}{\left[{\text{H}}_{\text{2}}\text{S}\right]} \\ {\text{K}}_{\text{a}}{\text{ = K}}_{a1}{\text{×k}}_{a2} \\ {\text{K}}_{\text{a}}{\text{ = 9×10}}^{-25} \end{gathered}$$

That quantity of needs to dissociate in the solution for [${\text{S}}^{\text{2 - }}$] to reach$1.1\times {10}^{-17}$ , thus we have:

$$\begin{gathered} {\text{K}}_{\text{a}}\text{ = }\frac{\left[{\text{S}}^{\text{2 - }}\right]{\left[{\text{H}}^{\text{ + }}\right]}^{\text{2}}}{\left[{\text{H}}_{\text{2}}\text{S}\right]} \\ {\text{K}}_{\text{a}}\text{ = }\frac{{\text{1.1×10}}^{\text{ - 17}}\text{×}{\left[{\text{H}}^{\text{ + }}\right]}^{\text{2}}}{{\text{0.1M - 1.1×10}}^{\text{ - 17}}} \end{gathered}$$

We can now quickly calculate [${\text{H}}^{\text{ + }}$] and pH:

$$\begin{gathered} {\left[{\text{H}}^{\text{ + }}\right]}^{\text{2}}{\text{ = K}}_{\text{a}}\text{×}\frac{{\text{0.1 - 1.1×10}}^{\text{ - 17}}}{{\text{1.1×10}}^{\text{ - 17}}} \\ {\text{ = 910}}^{\text{ - 25}}\text{×}\frac{{\text{0.1 - 1.1×10}}^{\text{ - 17}}}{{\text{1.1×10}}^{\text{ - 17}}} \\ {\text{ = 8.18×10}}^{\text{ - 9}} \\ {\text{[H}}^{\text{ + }}\text{]=}\sqrt{{\text{8.18×10}}^{\text{ - 9}}} \\ {\text{ = 9.04×10}}^{\text{ - 5}} \\ \text{pH = - log}\frac{{\text{[H}}^{\text{ + }}\text{]}}{\text{M}} \\ {\text{ = - log(9.04×10}}^{\text{ - 5}}\text{)} \\ \text{ = 4.04} \end{gathered}$$

Therefore, the value is:$\text{pH = 4.04}$ .