Question

A solution contains 0.274 M sodium hypochlorite and 0.146 M hypochlorous acid. What is the pH of the solution?

Answer 1

`7.73`

Explanation:

Your tool of choice for this problem will be the Henderson - Hasselbalch equation, which, for a buffer solution that consists of a weak acid and its conjugate base, allows you to calculate buffer pH by

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

In your case, the buffer consists of hypochlorous acid, `"HClO"`, a weak acid, and its conjugate base, the hypochlorite anion, `"ClO"^(-)`, its conjugate base.

The hypochlorite anions are deliver to the solution via sodium hypochlorite, `"NaClO"`, a salt of the hypochlorite ion.

`"NaClO"_text((aq]) -> "Na"_text((aq])^(+) + "ClO"_text((aq])^(-)`

The salt dissociates in a `1:1` ratio with the hypochlorite ion, which means that you have

`["NaClO"] = ["ClO"^(-)]`

You can find hypochlorous acid's `pK_a` here

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

So, before plugging in your values into the H - H equation, try to predict what you expect the result to be.

Notice that at equal concentrations of weak acid and conjugate base, the pH of the buffer is equal to the acid's `pK_a`.

In your case, the concentration of conjugate base is higher than the concentration of the weak acid, which means that the pH of the solution will be a little higher than the acid's `pK_a`.

So, the pH of the buffer will be equal to

`"pH" = 7.46 + log( (0.274color(red)(cancel(color(black)("M"))))/(0.146color(red)(cancel(color(black)("M"))))) = color(green)(7.73)`

Indeed, at these concentrations of weak acid and conjugate base, the buffer's pH is higher than the acid's `pK_a`.