Question

A sample of hard water contains $96\text{ppm}$ of $SO{4}_4^{2-}$ ions, $183\text{ppm}$ of $HC{O}_3^{-}$ ions and the only cation present is $C{a}^{2+}$. $X$ moles of $CaO$ are required to remove the $HC{O}_3^{-}$ ions from $1000\text{kg}$ of this sample of water. When the same amount of $CaO$ is added to the water sample, the concentration of residual $C{a}^{2+}$ ions (in ppm) becomes Y.
(Assume $CaC{O}_3$to be completely insoluble in water)
If the residual $C{a}^{2+}$ ions in $1\text{L}$ of the above -mentioned water sample is completely replaced by hydrogen ions, what will the $pH$ of the resultant sample of hard water?

• A. 2
• B. 2.7
• C. 3
• D. 3.3

Answer 1

The correct option is B 2.7

Density of hard water is approximately $1g/ml$
$1L=1000g=1kg$
$183$ ppm of $HC{O}_3^{-}$ represent 183 mg/L or $3\times {10}^{-3}$ moles per 1 kg, So for 1000 kg sample of hard water number of moles of $HC{O}_3^{-}$
$=3moles$
We know that ratio of $C{a}^{2+}$ to $HC{O}_3^{-}$ is $1:2$ in $Ca(HC{O}_3{)}_2$ , number of moles of $C{a}^{2+}=1.5$ moles.

Intially number of moles of $Ca(HC{O}_3{)}_2=1.5$
Complete balanced reaction to remove $Ca(HC{O}_3{)}_2$
$\underset{1}{Ca(HC{O}_3{)}_2}+\underset{1}{CaO}\to \underset{1}{CaC{O}_3}+H_{2}O+C{O}_2$

One mole of $Ca(HC{O}_3{)}_2$ replaced by one mole of $CaO$ so 1.5 mole of $CaO$required to remove 1.5 mole of $Ca(HC{O}_3{)}_2$.
And solution contains $S{O}_4^{2-}$ ions which counter ion is $C{a}^{2+}$ of molecular formula $CaS{O}_4$
As given:
$S{O}_4^{2-}=96ppm=1\times {10}^{-3}moles$
$1\times {10}^{-3}molesofC{a}^{2+}=40ppm$
All $Ca(HC{O}_3{)}_2$ so only $40ppm$ of $C{a}^{2+}$ ions left in solution.

One $C{a}^{2+}$ will replaced by 2 $H^{+}$ ions so for 1000 kg of sample-

$1moleofC{a}^{2+}replacedby$
$2moleof{H}^{+}$ or concentration of
$\left[H^{+}\right]=\frac{2}{1000L}=2\times {10}^{-3}$
$pH=-log(2\times {10}^{-3})=2.6989$