Question

Assuming that concentration of ${Ca}^{2+}$ ions in solution is in equal equivalence ratio to chloride ions, the hardness of water is :

• A. $9.180\times {10}^3$ ppm
• B. $6.180\times {10}^3$ ppm
• C. $12.180\times {10}^3$ ppm
• D. $4.180\times {10}^3$ ppm

Answer 1

Answer: (A)

$9.180\times {10}^3$ ppm

One drop of $AgN{O}_3=12.5\times {10}^{-3}$ g ${Cl}^{-}$
$\therefore 12$ drops $AgN{O}_3=12.5\times {10}^{-3}\times 12$ g ${Cl}^{-}$ $=150\times {10}^{-3}$ g ${Cl}^{-}$
So, $23$ mL sample $=0.15$ g ${Cl}^{-}$
$1000$ mL sample $=\frac{0.15\times 1000}{23}=6.52$ g in solution
Thus, molarity of ${Cl}^{-}$ in sample =$\frac{6.52}{35.5}=0.1837m$
Also given,
Eq. of $CaC{O}_3$ $=$ Eq. of ${Ca}^{2+}$ $=$ Eq. of ${Cl}^{-}$ (for $1$ litre water)
$\frac{w}{\frac{100}{2}}=0.1837$
$⟹$ $w_{CaC{O}_3}=0.1837\times 1002=9.18$ g
$\therefore$Hardness $=\frac{9.18\times {10}^6}{{10}^3}=9180$ ppm