Question

$100 \,mL$ 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 $20 \,mL$ of $0.025 \,M$ $NaOH$ for the titration. Thus, hardness of water expressed as ppm of $Ca ^{2+}$ ion is

• A. 100
• B. 50
• C. 40
• D. 20

Answer 1

Answer: (A)

$H _{2} R + Ca ^{2+} \longrightarrow Rca +2 H ^{+}$

$H ^{+}+ OH \longrightarrow H _{2} O$

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

Water coming out of cation exchange resin is acidic in nature and is neutralised by $OH^-$.

$N _{1} v _{1}= N _{2} V _{2}$

$( H ^{+}$ 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 $\left[ Ca ^{2+}\right]$

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

$\Rightarrow 5 \times 10^{-3} N =20 \times 5 \times 10^{-3} \,gL ^{-1} $

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

$=100 \,ppm \,Ca ^{2+}$