Q76P - Chemistry: Molecular Nature Of Matter And Change

Question

Wastewater from a cement factory contains 0.25 g of ${\mathbf{Ca}}^{\mathbf{2}\mathbf{+}}$ ion and 0.056 g of  ${\mathbf{Mg}}^{\mathbf{2}\mathbf{+}}$ ion per 100.0 L of solution. The solution density is 1.001 g/mL. Calculate the ${\mathbf{Ca}}^{\mathbf{2}\mathbf{+}}$ and ${\mathbf{Mg}}^{\mathbf{2}\mathbf{+}}$ concentrations in ppm (by mass).

Step-by-Step Solution

Answer

Hard water is a type of water that does not form lather with a detergent. The given wastewater from a cement factory contains 0.25 g of  $C{a}^{2+}$ and 0.056 g of  $M{g}^{2+}$ ions in per 100 L of solution and has a concentration of 2.5 ppm of $C{a}^{2+}$  and 0.56 ppm of  $M{g}^{2+}$.

 

Concentration of the Solution

A solution can be formed from the dissolution of a solute in a solvent. A solute decides the nature of a solution.

Parts by mass can be defined as the percentage of the ratio of the mass of a solute to the total mass of a solution.

 

 $\mathbf{Parts}\mathbf{ }\mathbf{by}\mathbf{ }\mathbf{Mass}\mathbf{=}\frac{\mathbf{Mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Solute}}{\mathbf{Mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{the}\mathbf{ }\mathbf{Solution}}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{.}$

Density can be defined as the ratio of the mass of a matter to the volume of the matter.

 $\mathbf{Density}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{Volume}}.$

Expression to calculate the concentration

Given information:

Waste water is also called hard water, and the given wastewater contains 0.25 g of  $C{a}^{2+}$ ion and 0.056 g of  $M{g}^{2+}$ ions in 100.0 L of a solution.

 

Volume of the solution $=100000\mathrm{mL}$

Density of the solution  $=1.001\mathrm{g}/\mathrm{mL}$

 

 $$\begin{gathered} \mathrm{De}nsity of the sol\mathrm{ution}=\frac{\mathrm{Mass}}{\mathrm{Volume}} \\ 1.001\mathrm{g}/\mathrm{mL}=\frac{\mathrm{Mass}}{100000 \mathrm{mL}} \\ \mathrm{Mass}=1.001\mathrm{g}/\mathrm{mL}\times 100000 \mathrm{mL} \\ =100100 \mathrm{g}. \end{gathered}$$

Firstly, the concentration of $C{a}^{2+}$  in hard water is:

The mass of the solute, ${\mathrm{Ca}}^{2+}=0.25\mathrm{g}$  

The molar mass of the solution $=100100\mathrm{g}$ 

 

 $$\begin{gathered} \mathrm{Parts} \mathrm{by} \mathrm{Mass}=\frac{\mathrm{Mass} \mathrm{of} \mathrm{Solute}}{\mathrm{Mass} \mathrm{of} \mathrm{the} \mathrm{Solution}}\times {10}^6 \\ \mathrm{Parts} \mathrm{by} \mathrm{Mass} \mathrm{of} {\mathrm{Ca}}^{2+}=\frac{0.25\mathrm{g}}{100100\mathrm{g}}\times {10}^6 \\ =2.50 \mathrm{ppm}. \end{gathered}$$

The concentration of the $M{g}^{2+}$ in the hard water is:

The mass of the solute, ${\mathrm{Mg}}^{2+}=0.056\mathrm{g}$

The molar mass of the solution $=100100g$

 $$\begin{gathered} \mathrm{Parts} \mathrm{by} \mathrm{Mass}=\frac{\mathrm{Mass} \mathrm{of} \mathrm{Solute}}{\mathrm{Mass} \mathrm{of} \mathrm{the} \mathrm{Solution}}\times {10}^6 \\ \mathrm{Parts} \mathrm{by} \mathrm{Mass} \mathrm{of} {\mathrm{Mg}}^{2+}=\frac{0.056\mathrm{g}}{100100\mathrm{g}}\times {10}^6 \\ =0.56 \mathrm{ppm}. \end{gathered}$$