The total concentration of \(\mathrm{Ca}^{2+}\) and \(\mathrm{Mg}^{2+}\) in a sample of hard water was determined by titrating a 0.100-L sample of the water with a solution of EDTA^{4-} \text { . The EDTA } ^ { 4 - } \text { chelates } the two cations: $$ \begin{aligned} \mathrm{Mg}^{2+}+[\mathrm{EDTA}]^{4-} & \longrightarrow[\mathrm{Mg}(\mathrm{EDTA})]^{2-} \\ \mathrm{Ca}^{2+}+[\mathrm{EDTA}]^{4-} & \longrightarrow[\mathrm{Ca}(\mathrm{EDTA})]^{2-} \end{aligned} $$ It requires \(31.5 \mathrm{~mL}\) of \(0.0104 \mathrm{M}\) [EDTA \(]^{4-}\) solution to reach the end point in the titration. A second 0.100 -L sample was then treated with sulfate ion to precipitate \(\mathrm{Ca}^{2+}\) as calcium sulfate. The \(\mathrm{Mg}^{2+}\) was then titrated with \(18.7 \mathrm{~mL}\) of 0.0104 \(M[\mathrm{EDTA}]^{4-}\). Calculate the concentrations of \(\mathrm{Mg}^{2+}\) and \(\mathrm{Ca}^{2+}\) in the hard water in \(\mathrm{mg} / \mathrm{L}\).

Step by step solution

01

Calculate moles of EDTA in the first and second titrations

To find out the amounts of \(\mathrm{Mg}^{2+}\) and \(\mathrm{Ca}^{2+}\) ions in the water, we first need to determine the moles of EDTA used in both titrations. We do this by multiplying the molarity of EDTA solution with the volume of EDTA used in each titration. For the first titration: Moles of EDTA = Molarity of EDTA * Volume of EDTA Moles of EDTA = (0.0104 mol/L) * (31.5 mL) * \(\frac{1 L}{1000 mL}\) For the second titration: Moles of EDTA = Molarity of EDTA * Volume of EDTA Moles of EDTA = (0.0104 mol/L) * (18.7 mL) * \(\frac{1 L}{1000 mL}\)

02

Calculate moles of \(\mathrm{Mg}^{2+}\) and \(\mathrm{Ca}^{2+}\)

Since each mole of EDTA in the first titration chelates one mole of either \(\mathrm{Mg}^{2+}\) or \(\mathrm{Ca}^{2+}\), the total moles of metal ions = moles of EDTA in the first titration. The moles of EDTA required in the second titration are equal to moles of \(\mathrm{Mg}^{2+}\) ions because after the precipitation of the \(\mathrm{Ca}^{2+}\) ions, only \(\mathrm{Mg}^{2+}\) ions remain in the solution. Total moles of metal ions (first titration) = moles of EDTA in the first titration Moles of \(\mathrm{Mg}^{2+}\) (second titration) = moles of EDTA in the second titration Then to find the moles of \(\mathrm{Ca}^{2+}\) ions, subtract the moles of \(\mathrm{Mg}^{2+}\) from the total moles of metal ions: Moles of \(\mathrm{Ca}^{2+}\) = Total moles of metal ions - Moles of \(\mathrm{Mg}^{2+}\)

03

Calculate concentrations of \(\mathrm{Mg}^{2+}\) and \(\mathrm{Ca}^{2+}\)

Now that we have the moles of \(\mathrm{Mg}^{2+}\) and \(\mathrm{Ca}^{2+}\) ions, we can calculate their concentrations in the original water sample by dividing the moles by the volume of the water sample (given as 0.100 L): Concentration of \(\mathrm{Mg}^{2+}\) = \(\frac{\text{moles of }\mathrm{Mg}^{2+}}{\text{volume of water sample}}\) Concentration of \(\mathrm{Ca}^{2+}\) = \(\frac{\text{moles of }\mathrm{Ca}^{2+}}{\text{volume of water sample}}\)

04

Convert concentrations to mg/L

To express the concentrations of \(\mathrm{Mg}^{2+}\) and \(\mathrm{Ca}^{2+}\) in mg/L, multiply the concentrations in mol/L by their respective molar masses (24.305 g/mol for \(\mathrm{Mg}^{2+}\) and 40.078 g/mol for \(\mathrm{Ca}^{2+}\)) and then by 1000 to convert to mg/L: Concentration of \(\mathrm{Mg}^{2+}\) in mg/L = Concentration of \(\mathrm{Mg}^{2+}\) in mol/L * Molar mass of \(\mathrm{Mg}^{2+}\) * 1000 Concentration of \(\mathrm{Ca}^{2+}\) in mg/L = Concentration of \(\mathrm{Ca}^{2+}\) in mol/L * Molar mass of \(\mathrm{Ca}^{2+}\) * 1000

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