# What is the "pH" of a "0.49-M" "NaClO" solution?

Apr 3, 2018

$\text{pH} = 14 + \log \left(\sqrt{0.49 \cdot {K}_{b}}\right)$

#### Explanation:

Sodium hypochlorite is soluble in water, so it dissociates completely in aqueous solution to produce sodium cations, ${\text{Na}}^{+}$, and hypochlorite anions, ${\text{ClO}}^{-}$, in $1 : 1$ mole ratios.

This means that you have

["ClO"^(-)]_ 0 = ["NaClO"] = "0.49 M"

The hypochlorite anions will act as a weak base in aqueous solution, meaning that some of these anions will react with water to form hypochlorous acid, $\text{HClO}$, and hydroxide anions, ${\text{OH}}^{-}$.

${\text{HClO"_ ((aq))^(-) + "H"_ 2"O"_ ((l)) rightleftharpoons "HClO"_ ((aq)) + "OH}}_{\left(a q\right)}^{-}$

So right from the start, you should expect the $\text{pH}$ of the solution to be $> 7$ because of the presence of the hypochlorite anions.

Now, the base dissociation constant of the hypochlorite anions is calculated using the equation

${K}_{b} = \left(\left[{\text{HClO"] * ["OH"^(-)])/(["ClO}}^{-}\right]\right)$

So if you start with $\text{0.49 M}$ of hypochlorite anions and have $x$ $\text{M}$ ionize to produce hypochlorous acid and hydroxide anions, then the $1 : 1$ mole ratios present in the balanced chemical equation that describes the ionization equilibrium tell you that, at equilibrium, you have

["HClO"] = ["OH"^(-)] = x quad "M"

If $x$ $\text{M}$ of hypochlorite anions ionize, then you get $x$ $\text{M}$ of hypochlorous acid and $x$ $\text{M}$ of hydroxide anions.

["ClO"^(-)] = (0.49 - x) quad "M"

If $x$ $\text{M}$ of hypochlorite anions ionize, then you are left with $\left(0.49 - x\right)$ $\text{M}$ of hypochlorite anions.

The expression of the base dissociation constant will be

${K}_{b} = \frac{x \cdot x}{0.49 - x}$

${K}_{b} = {x}^{2} / \left(0.49 - x\right)$

Now, you didn't provide a value for the base dissociation constant, but I can tell you that it's small enough to allow you to use the approximation

$0.49 - x \approx 0.49$

This means that you have

${K}_{b} = {x}^{2} / 0.49$

and so

$x = \sqrt{0.49 \cdot {K}_{b}}$

Since we've said that $x$ $\text{M}$ represents the equilibrium concentration of hydroxide anions, you will have

["OH"^(-)] = sqrt(0.49 * K_b) quad "M"

As you know, an aqueous solution at ${25}^{\circ} \text{C}$ has

$\text{pH + pOH = 14}$

Since

"pOH" = - log(["OH"^(-)])

you can say that the $\text{pH}$ of a solution is given by

"pH" = 14 + log(["OH"^(-)])

In your case, this will be equal to

$\text{pH} = 14 + \log \left(\sqrt{0.49 \cdot {K}_{b}}\right)$

Now all you have to do is to use the value of the base dissociation constant given to you--or you can simply do a quick search for the base dissociation constant of the hypochlorite anion--and plug it into the equation to find the $\text{pH}$ of the solution.

The answer should be rounded to two decimal places because you have two sig figs for the concentration of the solution.

Alternatively, you can look up the acid dissociation constant, ${K}_{a}$, of hypochlorous acid and use the fact that at ${25}^{\circ} \text{C}$, you have

${K}_{a} \cdot {K}_{b} = 1.0 \cdot {10}^{- 14} \implies {K}_{b} = \frac{1.0 \cdot {10}^{- 14}}{K} _ a$

This will get you

$\text{pH} = 14 + \log \left(\sqrt{0.49 \cdot \frac{1.0 \cdot {10}^{- 14}}{K} _ a}\right)$