A water sample was tested for water hardness a 5000 ml...

A water sample was tested for water hardness. A 50.00 mL sample, prepared and buffered to pH 10, required 23.55 mL of 0.0250 M EDTA. A second 50.00 mL aliquot, prepared with NaOH, was titrated with 14.53 mL of the same EDTA solution. a. Calculate the total hardness of the water sample, express in ppm CaCO3 b. Calculate the concentration of CaCO3 in the sample in ppm c. Calculate the concentration of MgCO3 in the sample in ppm

The total concentration of $\mathrm{Ca}^{2+}$ and $\mathrm{Mg}^{2+}$ in a sample of hard water was determined by titrating a $0.100-\mathrm{L}$ sample of the water with a solution of EDTA $^{4-}$. The EDTA $^{4-}$ 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 $0.0104 \mathrm{M}[\mathrm{EDTA}]^{4-}$ solution to reach the end point in the titration. A second $0.100-\mathrm{L}$ 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 $0.0104 \mathrm{M}$ $[\mathrm{EDTA}]^{4-},$ Calculate the concentrations of $\mathrm{Mg}^{2+}$ and $\mathrm{Ca}^{2+}$ in the hard water in $\mathrm{mg} / \mathrm{L}$.

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