Question

$100mL$ of sample of hard water is passed through a column of the cation exchange resin $H_{2}R$. The water coming out of the column requires $20mL$ of $0.025M$ $NaOH$ for the titration. Thus, hardness of water expressed as ppm of $C{a}^{2+}$ ion is

• A. 100
• B. 50
• C. 40
• D. 20

Answer 1

Answer: (A)

$H_{2}R+C{a}^{2+}⟶Rca+2H^{+}$

$H^{+}+OH⟶H_{2}O$

Thus, $\left[H^{+}\right]\equiv [OH]\equiv \left[C{a}^{2+}\right]$ in terms of equivalent.

Water coming out of cation exchange resin is acidic in nature and is neutralised by $O{H}^{-}$.

$N_1{v}_1=N_2{V}_2$

$(H^{+}$ in water $)=NaOH$

$N_1\times 100=0.025\times 20$

$N_1=\frac{0.025\times 20}{100}$

$=5\times 10^{-3}N$

Equivalent of $\left[H^{+}\right]=$ Equivalent of $\left[C{a}^{2+}\right]$

$\therefore \left[C{a}^{2+}\right]=\left[H^{+}\right]$

$\Rightarrow 5\times 10^{-3}N=20\times 5\times 10^{-3}g{L}^{-1}$

$=\frac{100\times 10^{-3}}{10^3}\times 10^{6}g$ in $10^{6}mL$

$=100ppmC{a}^{2+}$