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-} \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 \(\mathrm{mg} / \mathrm{L}\).
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.