Question

A sample of water (pH = 10) contains 45 mg/l of $C{O}_3^{-}$ ion.
$HC{O}_3^{-}\stackrel{k}{\rightleftharpoons }{H}^{+}+C{O}_3^{-}$
The value of k is $4.5\times {10}^{-11}$ mole/litre. What is the total alkalinity of water in mg/l as $CaC{O}_3$?

• A. 75
• B. 83.5
• C. 158.5
• D. 163.5

Answer 1

The correct option is D 163.5

$\left[C{O}_3^{-}\right]=45\frac{mg}{l}=\frac{45}{60}\times {10}^{-3}$

$=7.5\times {10}^{-4}mol/l=1.5meq/lit$

$\left[H^{+}\right]={10}^{-10}mol/litres=1\times {10}^{-7}meq/lit$

Also, $\left[O{H}^{-}\right]={10}^{-4}mol/lit=0.1\text{meq/lit}.$

$\left[HC{O}_3^{-}\right]=\frac{\left[H^{+}\right]\left[C{O}_3^{-}\right]}{k}$

$=\frac{{10}^{-10}\times 7.5\times {10}^{-4}}{4.5\times {10}^{-11}}=1.67\times {10}^{-3}\frac{mol}{litre}$

$=1.67\times {10}^{-3}g-eq/l=1.67\text{meq/lit}.$

Total alkalinity $=\left[HC{O}_3^{-}\right]+\left[C{O}_3^{2-}\right]+\left[O{H}^{-}\right]-\left[H^{+}\right]$

$=(1.67+1.5+0.1-1\times {10}^{-7})\text{meq/lit}$

= 3.27 meq/lit

$=(3.27\times 50)\text{mg/lit}$ as $CaC{O}_3$

$=163.5\text{mg/lit}$ as $CaC{O}_3$