Question

500.0 mL of 0.140 M NaOH is added to 565 mL of 0.250 M weak acid (Ka = 2.92 × 10-5). What is the pH of the resulting buffer?

Answer 1

`"pH" = 4.528`

Explanation:

Right from the start, looking at the values given to you, I'd say that the `"pH"` is very close to the `"p"K_a` of the weak acid.

So I would say that you have

`"pH" ~~ "p"K_a`

or

`"pH" ~~ - log(2.92 * 10^(-5))`

`"pH" ~~ 4.535`

Here's why.

The idea here is that the hydroxide anions delivered to the solution by the soluble salt will react with the weak acid, let's say `"HA"`, to form the conjugate base of the acid, `"A"^(-)`.

`"HA"_ ((aq)) + "OH"_ ((aq))^(-) -> "A"_ ((aq))^(-) + "H"_ 2"O"_ ((l))`

The weak acid and the hydroxide anions react in a `1:1` mole ratio to produce conjugate base. In other words, every mole of hydroxide anions added to the solution will consume `1` mole of weak acid and produce `1` mole of conjugate base.

Use the molarities and the volumes of the two solutions to figure out how many moles of each reactant are being mixed

`500.0 color(red)(cancel(color(black)("mL"))) * "0.140 moles OH"^(-)/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.0700 moles OH"^(-)`

`565 color(red)(cancel(color(black)("mL"))) * "0.250 moles HA"/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.141 moles HA"`

Now, you have fewer moles of hydroxide anions than moles of weak acid, which implies that the hydroxide anions will act as a limiting reagent, i.e. they will be completely consumed by the reaction.

The resulting solution will thus contain

`n_ ("OH"^(-)) = "0 moles OH"^(-) ->` completely consumed

`n_ ("HA") = "0.141 moles" - "0.0700 moles" = "0.0710 moles HA"`

`n_ ("A"^(-)) = "0 moles" + "0.0700 moles" = "0.0700 moles A"^(-)`

The total volume of the solution will be

`V_"total" = "500.0 mL" + "565 mL" = "1065 mL"`

The concentration of the weak acid and of the conjugate base in the resulting solution will be

`["HA"] = "0.0710 moles"/(1065 * 10^(-3)"L") = "0.0667 M"`

`["A"^(-)] = "0.0700 moles"/(1065 * 10^(-3)"L") = "0.0657 M"`

Now, you can calculate the `"pH"` of the buffer by using the Henderson - Hasselbalch equation, which for your buffer will look like this

`"pH" = "p"K_a + log( (["A"^(-)])/(["HA"]))`

Plug in your values to find

`"pH" = 4.53 + log(( 0.0657 color(red)(cancel(color(black)("M"))))/(0.0667 color(red)(cancel(color(black)("M")))))`

`"pH" = 4.535 + (-0.00656)`

`color(darkgreen)(ul(color(black)("pH" = 4.528)))`

The answer is rounded to three decimal places, the number of sig figs you have for all but one of your values.

Notice that the initial prediction turned out to be correct, we really do have

`"pH" ~~ "p"K_a`

The prediction was based on the fact that your values showed that you had almost twice as many moles of weak acid than moles of strong base.

The `1:1` mole ratio that the two reactants share means that almost half of the number of moles of weak acid will be converted to moles of conjugate base.

So the concentration of the weak acid and the concentration of the conjugate base were going to come out as being almost equal.

And since the Henderson - Hasselbalch equation shows that when

`["HA"] = ["A"^(-)]`

you have

`"pH" = "p"K_a + log(1)`

`"pH" = "p"K_a`

the prediction was right on the money.