Question

Chlorine gas ($8mg/LasC{l}_2$) was added to a drinking water sample. If the free chlorine residual and pH was measured to be $2mg/L$ (as $C{l}_2$ ) and $7.5$, respectively, what is the concentration of residual $OC{l}^{-}$ ions in the water? Assume that the chlorine gas added to the water is completely converted to $HOCl$ and $OC{l}^{-}$. Atomic weight of Cl = 35.5
Given,
$OC{l}^{-}+H^{+}\stackrel{k}{\rightleftharpoons }HOClK={10}^{7.5}$

• A. $5.634\times {10}^{-5}mol/L$
• B. $2.817\times {10}^{-3}mol/L$
• C. $1.408\times {10}^{-5}mol/L$
• D. $1.127\times {10}^{-4}mol/L$

Answer 1

The correct option is C $1.408\times {10}^{-5}mol/L$

$pH=7.5$

$\left[H^{+}\right]={10}^{-pH}$

$={10}^{-7.5}$

Free residual chlorine

$(HOCl+OC{l}^{-})$

$=2mg/L$ as $C{l}_2$

$=2\times {10}^{-3}g/L$ as $C{l}_2$

$=\frac{2\times {10}^{-3}}{\text{molecular weight of}C{l}_2}mol/L$

$\left[HOCl\right]+\left[OC{l}^{-}\right]=\frac{2\times {10}^{-3}}{71}mol/L$

$\Rightarrow \left[HOCl\right]=\frac{2\times {10}^{-3}}{71}-\left[OC{l}^{-}\right]$

Given,
$OC{l}^{-}+H^{+}\stackrel{k}{\rightleftharpoons }HOCl$

$K={10}^{7.5}$

$K=\frac{\left[HOCl\right]}{\left[H^{+}\right]\left[OC{l}^{-}\right]}$

$\Rightarrow {10}^{7.5}=\frac{\frac{2\times {10}^{-3}}{71}\left[OC{l}^{-}\right]}{{10}^{-7.5}\left[OC{l}^{-}\right]}$

$2\times \left[OC{l}^{-}\right]=\frac{2\times {10}^{-3}}{71}$

$\Rightarrow \left[OC{l}^{-}\right]=1.408\times {10}^{-5}mol/L$