Chapter 23: Problem 98
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 \(^{4-} .\) The EDTA \(^{4-}\) chelatesthe two cations: $$\begin{array}{c}{\mathrm{Mg}^{2+}+[\mathrm{EDTA}]^{4-} \longrightarrow[\mathrm{Mg}(\mathrm{EDTA})]^{2-}} \\\ {\mathrm{Ca}^{2+}+[\mathrm{EDTA}]^{4-} \longrightarrow[\mathrm{Ca}(\mathrm{EDTA})]^{2-}}\end{array}$$ 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-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 \(\mathrm{mg} / \mathrm{L} .\)
Short Answer
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