Question

A sample of hard water contains 96 ppm of $S{{O}_{4}}^{2-}$ and 183 ppm of $HC{{O}_{3}}^{-}$ with $C{{a}^{2+}}$ as the only cation. How many moles of CaO will be required to remove $HC{{O}_{3}}^{-}$ from 1000 kg of this water ? If 1000 kg of this water is treated with the amount of CaO calculated above, what will be the concentration (in ppm) of residual $C{{a}^{2+}}$ions? (Assume $CaC{{O}_{3}}$ to be completely insoluble in water). If the $C{{a}^{2+}}$ ions in one litre of the treated water are completely exchanged with hydrogen ions, what will be its pH? (one ppm means one part of the substance in one million part of water).

Answer 1

in 1000kg of water 1ppm$=0.001\times 1000=1g$
Use formula: $\text{no}\text{. of moles = }\dfrac{\text{given mass}}{\text{molar mass}}$to calculate number of moles of $S{{O}_{4}}^{2-}$and $HC{{O}_{3}}^{-}$ions and calculate the mass of calcium ions.
The addition of CaO to 1000kg of hard water causes the following reaction-
$CaO+Ca{{(HC{{O}_{3}})}_{2}}\to 2CaC{{O}_{3}}+{{H}_{2}}O$
Concentration in ppm $=\dfrac{mass\text{ of C}{{\text{a}}^{2+}}}{total\text{ mass of solution}}\times {{10}^{6}}$
$\text{pH= -}\log ({{H}^{+}})$

Complete step by step solution:
we know that 1ppm $=.001g/kg$ that is, 0.001g in 1 kg.
Therefore, in 1000kg of water 1ppm$=0.001\times 1000=1g$
Similarly, in 1000kg of water, 96 ppm of $S{{O}_{4}}^{2-}=96g$
And 183 ppm of $HC{{O}_{3}}^{-}=183g$
We know that $\text{no}\text{. of moles = }\dfrac{\text{given mass}}{\text{molar mass}}$
No. of moles of $S{{O}_{4}}^{2-}=\dfrac{96}{96}=1mol$
No. of moles of $HC{{O}_{3}}^{-}=\dfrac{183}{61}=3moles$
$S{{O}_{4}}^{2-}$ and $HC{{O}_{3}}^{-}$ ions are present as $CaS{{O}_{4}}$ and $Ca{{(HC{{O}_{3}})}_{2}}$.
Consequently, amount of $C{{a}^{2+}}$ ions present $=1+\dfrac{3}{2}=2.5mol$
The addition of CaO to 1000kg of hard water causes the following reaction-
$CaO+Ca{{(HC{{O}_{3}})}_{2}}\to 2CaC{{O}_{3}}+{{H}_{2}}O$
As we can see from the above equation that 1 mol of CaO can remove 1 mol of$Ca{{(HC{{O}_{3}})}_{2}}$. On this basis, 1.5 mole of CaO can remove 1.5 mole of $Ca{{(HC{{O}_{3}})}_{2}}$ in form of $CaC{{O}_{3}}$.
NOW, in the treated water, only $CaS{{O}_{4}}$ is left. Therefore, 1 mol of $C{{a}^{2+}}$ ions will be present in 1000kg of water.
We know that $\text{no}\text{. of moles = }\dfrac{\text{given mass}}{\text{molar mass}}$
Therefore, mass of 1 mol of $C{{a}^{2+}}$ ions$=molar\text{ mass of C}{{\text{a}}^{2+}}\times no.\text{ of moles of C}{{\text{a}}^{2+}}=40g/mol\times 1mol=40g$ …(i)
Concentration in ppm $=\dfrac{mass\text{ of C}{{\text{a}}^{2+}}}{total\text{ mass of solution}}\times {{10}^{6}}$
Concentration (in ppm) of residual $C{{a}^{2+}}$ions$=\dfrac{40g}{{{10}^{6}}g}\times {{10}^{6}}=40ppm$
As we just saw, mass of $C{{a}^{2+}}$ ions in 1000kg (1000L) of water is 40 g [from equation (i)], now the mass $C{{a}^{2+}}$ ions in 1 litres would be $40\times {{10}^{-3}}g$ as $1kg=1L$ .
We know that $Molarity=\dfrac{given\text{ mass}}{\text{molar mass}}\times \dfrac{1}{volum{{e}_{Litres}}}$
Molarity of $C{{a}^{2+}}$ $=\dfrac{40\times {{10}^{-3}}g}{40g/mol}\times \dfrac{1}{1L}={{10}^{-3}}mol/L$
According to the question, if the $C{{a}^{2+}}$ ions in one litre of the treated water are completely exchanged with hydrogen ions than molarity of hydrogen ions would be

& [{{H}^{+}}]=2\times {{10}^{-3}}mol/L \\\ & and\text{ pH= -}\log (2\times {{10}^{-3}})=2.7 \\\ \end{aligned}$$ **Therefore, if the $C{{a}^{2+}}$ ions in one litre of the treated water are completely exchanged with hydrogen ions, then the pH would be 2.7.** **Note:** Molarity, ppm, molality, all these are concentration terms. Molarity should not be confused with molality. Molality(m) is defined as the number of moles of a solute per kilogram of a solvent whereas Molarity(M) is defined as the number of moles of a solute added in a solution per liter of solution.