# Question #cda49

May 10, 2015

The pH of the solution will be equal to 10.1.

Sodium hypochlorite will dissociate in aqueous solution to give sodium cations, $N {a}^{+}$, and hypochlorite anions, $C l {O}^{-}$.

The hypochlorite anions will act as a base and react with the water to form hypochlorous acid, $H C l O$, and hydroxide ions, $O {H}^{-}$. You're going to have to determine the base dissociation constant, ${K}_{b}$, for the hypochlorite ion.

${K}_{b} = {K}_{w} / {K}_{a} = {10}^{- 14} / \left(4.0 \cdot {10}^{- 8}\right) = 2.5 \cdot {10}^{- 7}$

Now use an ICE table to solve for the concentration of hydroxide ions.

$\text{ } C l {O}_{\left(a q\right)}^{-} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s H C l {O}_{\left(a q\right)} + O {H}_{\left(a q\right)}^{-}$
I......0.081......................................0.................0
C.......(-x)......................................(+x)..............(+x)
E....0.081-x....................................x.................x

By definition, the base dissociation constant will be

${K}_{b} = \frac{\left[H C l O\right] \cdot \left[O {H}^{-}\right]}{\left[C l {O}^{-}\right]} = \frac{x \cdot x}{0.081 - x} = {x}^{2} / \left(0.081 - x\right)$

Because the base dissociation constant is so small, you can approximate (0.081 - x) with 0.081. This will get you

${K}_{b} = {x}^{2} / 0.081 = 2.5 \cdot {10}^{- 7} \implies x = 1.42 \cdot {10}^{- 4}$

This will be equal to the concentration of the hydroxide ions

$\left[O {H}^{-}\right] = 1.42 \cdot {10}^{- 4} \text{M}$

The solution's pOH will be

$p O H = - \log \left(\left[O {H}^{-}\right]\right)$

$p O H = - \log \left(1.42 \cdot {10}^{- 4}\right) = 3.9$

Thus, the solution's pH will be

$p {H}_{\text{sol}} = 14 - p O H = 14 - 3.9 = \textcolor{g r e e n}{10.1}$