Question

If 30 mL of 0.10 M NaOH is added to 40 mL of 0.20 M HC2H3O2, what is the pH of the resulting solution at 25°C? Ka for HC2H3O2 is 1.8 x 10^–5 at 25°C.

Answer 1

See below:

Explanation:

The reaction that will occur is:

`NaOH(aq)+CH_3COOH(aq) -> CH_3COONa + H_2O(l)`

Now, using the concentration formula we can find the amount of moles of `NaOH` and Acetic acid:

`c=(n)/v`

For `NaOH`
Remember that `v` should be in litres, so divide any milliliter values by 1000.

`cv=n`

`0.1 times 0.03=0.003 mol` of `NaOH`

For `CH_3COOH`:

`cv=n`

`0.2 times 0.04=0.008 mol` of `CH_3COOH`.

So 0.003 mol of `NaOH` will react to completion with the acid to form 0.003 mol of Sodium acetate, `CH_3COONa`, in the solution, along with 0.005 mol of acid dissolved in a total volume of 70 ml. This will create an acidic buffer solution.

Let's find the concentration of the salt and the acid, respectively:

`c_(acid)=(0.005)/0.7 approx 0.0714 mol dm^-3`

`c_(sa l t)=(0.003)/0.007 approx 0.0428 mol dm^-3`

Now, we can use theHenderson-Hasselbalch equation to find the `pH` of the resulting solution.

The equation looks like this:

`pH=pKa+log_10(([S a l t])/([Acid]))`

We are given the `K_a` of the acid, so the `pKa` is the negative logarithm of the `K_a` value.

`pKa=-log_10[K_a]`
`pKa=-log_10[1.8 times 10^-5]`
`pKa approx 4.74`

Now we have to plug in all the values into the equation:

`pH=4.74+log_10(([0.0428])/([0.0714]))`

`pH=4.74-0.2218`

`pH approx 4.52`