Question

Question #00924

Answer 1

`"pH" = 10.04`

Explanation:

Your tool of choice here will be the Henderson - Hasselbalch equation, which for a weak acid/conjugate base buffer looks like this

`color(blue)(ul(color(black)("pH" = - "p"K_a + log( (["conjugate base"])/(["weak acid"])))))`

In your case, the buffer contains phenol, `"C"_6"H"_5"OH"`, as the weak acid and the phenolate anion, `"C"_6"H"_5"O"^(-)`, as the conjugate base.

The conjugate base is delivered to the solution by sodium phenolate, a soluble salt.

In your case, the Henderson - Hasselbalch equation will be

`"pH" = "p"K_a + log ( (["C"_6"H"_5"O"^(-)])/(["C"_6"H"_5"OH"]))`

Notice that the buffer contains more conjugate base than weak acid. This should let you know that the pH of the solution will come out to be higher than the `"p"K_a` of the acid.

So, plug in your values to find

`"pH" = 10.00 + log ((1.2 color(red)(cancel(color(black)("mol L"^(-1)))))/(1.1color(red)(cancel(color(black)("mol L"^(-1))))))`

`color(darkgreen)(ul(color(black)("pH" = 10.04)))`

The answer is rounded to two decimal places, the number of sig figs you have for the molarities of the weak acid and of the conjugate base.

As predicted, the pH of the buffer is indeed higher, if only slightly, than the `"p"K_a` of the acid.