Question

$\text{Alkalinity of water, in equivalent/litre (eq/litre), is given by}\left\{HC{O}_3^{-}\right\}+2\left\{C{O}_3^{-2}\right\}+\left\{O{H}^{-}\right\}-\left\{H^{+}\right\}$
where, { } represents concentration in mol/litre. For a water sample , the concentrations of $HC{O}_3^{-}=2\times {10}^{-3}mol/litre,$ $C{O}_3^{2-}=3.04\times {10}^{-4}mol/litre$
and the pH of water =9.0. The atomic weights are: Ca=40; C=12 and O=16. If the concentration of $O{H}^{-}$and $H^{+}$ are $\text{NEGLECTED, the alkalinity of the water sample (in mg/litre as}CaC{O}_3$) is

• A. $50.0$
• B. $65.2$
• C. $100.0$
• D. $130.4$

Answer 1

The correct option is D $130.4$

Alkalinity of water (eq/lt)
$=\left\{HC{O}_3^{-}\right\}+2\left\{C{O}_3^{-2}\right\}+\left\{O{H}^{-}\right\}-\left\{H^{+}\right\}$

{ } $\to$ concentration in mol/lt

Given,
$HC{O}_3^{-}=2\times {10}^{-3}mol/lt$
$C{O}_3^{2-}=3.04\times {10}^{-4}mol/lt$

As per question,
On neglecting $H^{+}$ and $O{H}^{-}$ concentration
Alkalinity of water $=(2\times {10}^{-3}+2\times 3.04\times {10}^{-4})$

$=2.608\times {10}^{-3}geq/lt$

$=(2.608\times {10}^{-3}\times 50)gm/ltasCaC{O}_3$

$=130.4mg/ltasCaC{O}_3$