# How would you use the Henderson-Hasselbalch equation to calculate the pH of a solution that is 0.120 M in HClO and 0.185 M in KClO?

Nov 22, 2015

$\text{pH} = 7.65$

#### Explanation:

The Henderson - Hasselbalch equation allows you to calculate the pH of buffer solution that contains a weak acid and its conjugate base by using the concentrations of these two species and the $p {K}_{a}$ of the weak acid.

color(blue)("pH" = pK_a + log( (["conjugate base"])/(["weak acid"])))

In your case, the weak acid is hypochlorous acid, $\text{HClO}$. Its conjugate base, the hypochlorite anion, ${\text{ClO}}^{-}$, is delivered to the solution by one of its salts, potassium hypochlorite, $\text{KClO}$.

The acid dissociation constant, ${K}_{a}$, for hypochlorous acid is equal to $3.5 \cdot {10}^{- 8}$

http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf

Now, before doing any calculations, try to predict what you expect the solution's pH to be compared with the acid's $p {K}_{a}$.

Notice that when you have equal concentrations of weak acid and conjugate base, the log term will be equal to zero, since

$\log \left(1\right) = 0$

This tells you that if you have more conjugate base than weak acid, the log term will be greater than $1$, which will cause the pH to be higher than the $p {K}_{a}$.

With this in mind, plug in your values into the H-H equation to get

"pH" = -log(3.5 * 10^(-8)) + log( (0.185color(red)(cancel(color(black)("M"))))/(0.120color(red)(cancel(color(black)("M")))))

$\text{pH} = 7.46 + 0.188 = \textcolor{g r e e n}{7.65}$