Question

Wastewater from a cement factory contains 0.280 g of Ca2+ ion and 0.0220 g of Mg2+ ion per 100.0 L of solution. The solution density is 1.001 g/mL. Calculate the Ca2+ and Mg2+ concentrations in ppm (by mass).

Answer 1

Answer: The concentration of $Ca^{2+}\text{ and }Mg^{2+}$ ions are 2.797 ppm and 0.212 ppm respectively.

Explanation:

To calculate the mass of solution, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Volume of gold = 100 L = 100000 mL    (Conversion factor:  1 L = 1000 mL)

Density of gold = 1.001 g/mL

Putting values in above equation, we get:

$1.001g/mL=\frac{\text{Mass of solution}}{100000mL}\\\\\text{Mass of solution}=1.001\times 10^5g$

To calculate the concentration in ppm (by mass), we use the equation:

$ppm=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^6$

• Calculating the concentration of calcium ions:

Mass of $Ca^{2+$ ions = 0.280 g

Putting values in above equation, we get:

$ppm(Ca^{2+})=\frac{0.280g}{1.001\times 10^5}\times 10^6=2.797ppm$

• Calculating the concentration of magnesium ions:

Mass of $Mg^{2+$ ions = 0.0220 g

Putting values in above equation, we get:

$ppm(Mg^{2+})=\frac{0.0220g}{1.001\times 10^5}\times 10^6=0.212ppm$

Hence, the concentration of $Ca^{2+}\text{ and }Mg^{2+}$ ions are 2.797 ppm and 0.212 ppm respectively.