\(\mathrm{RH}_2\) (ion exchange resin) can replace \(\mathrm{Ca}^{2+}\) ions in hard water as:…
\(\mathrm{RH}_2\) (ion exchange resin) can replace \(\mathrm{Ca}^{2+}\) ions in hard water as:
\(\mathrm{RH}_2+\mathrm{Ca}^{2+} ightarrow \mathrm{RCa}+2 \mathrm{H}^{+}\)
If 1 L of hard water after passing through \(\mathrm{RH}_2\) has \(\mathrm{pH}=3\), then hardness in parts per million of \(\mathrm{Ca}^{2+}\) is:
10
100
20
40
Solution
Given reaction:
\(\mathrm{RH}_2+\mathrm{Ca}^{2+} ightarrow \mathrm{RCa}+2 \mathrm{H}^{+}\)
This shows that 1 mole of \(\mathrm{Ca}^{2+}\) releases 2 moles of \(\mathrm{H}^{+}\)upon ion exchange.
Calculate \(\left[\mathrm{H}^{+}ight]\)from pH
\(\mathrm{pH}=3 \Rightarrow\left[\mathrm{H}^{+}ight]=10^{-3} \mathrm{~mol} / \mathrm{L}\)
Use stoichiometry to find [ \(\mathrm{Ca}^{2+}\) ]
From the balanced reaction:
\(\left[\mathrm{Ca}^{2+}ight]=\frac{\left[\mathrm{H}^{+}ight]}{2}=\frac{10^{-3}}{2} \mathrm{~mol} / \mathrm{L}\)
Convert to ppm using molar mass of \(\mathrm{Ca}=\mathbf{4 0} \mathrm{g} / \mathrm{mol}\)
\(\operatorname{ppm}=\left(\frac{40 \times \frac{10^{-3}}{2}}{1}ight) \times 10^6=20 \mathrm{ppm}\)
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