Question

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?

Answer 1

`"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 `pK_a` of the weak acid.

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

In your case, the weak acid is hypochlorous acid, `"HClO"`. Its conjugate base, the hypochlorite anion, `"ClO"^(-)`, is delivered to the solution by one of its salts, potassium hypochlorite, `"KClO"`.

The acid dissociation constant, `K_a`, for hypochlorous acid is equal to `3.5 * 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 `pK_a`.

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

`log(1) = 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 `pK_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")))))`

`"pH" = 7.46 + 0.188 = color(green)(7.65)`