Question

Question #45d65

Answer 1

Here's how you sketch the titration curve.

Explanation:

Assume that you are titrating 50 mL of `"0.1 mol/L HA"` (`K_"a" = 10^-5`) with `"0.1 mol/L NaOH"`.

You know that the equivalence point will be at `"50 mL of NaOH"`. So start by drawing the pH and volume axes.

The vertical axis will have `"pH"` running from 0 to 14.

The horizontal axis will run from 0 mL to somewhere past 50 mL (say, 60 to 80 mL).

The equivalence point

You already know one point. The equivalence point has `pH 9` at 50 mL. So plot a point at (`50, 9`).

At half-equivalence

You know that at half-equivalence, `"pH" = "p"K_"a"`. Since `"p"K_"a" = 5`, plot a point at (`25, 5`).

At the start

If the acid had been 0.1 mol/L HCl, we would have the starting `"p"H = 1`. But this is a weak acid. It does not dissociate completely, so the starting `"pH"` will be higher, probably about `"pH 3"`. Draw a point at (`"0, 3"`).

Just after the start

With a weak acid, the pH rises rapidly at the beginning of the titration, perhaps by 1 pH unit in the first 10 % of the titration. Plot a point at (`5, 4`).

Just before the equivalence point

The `"pH"` starts to rise rapidly from `"pH 6"` at about 90 % of the titration. Plot a point at (`45,6`).

Well past the equivalence point

The `"pH"` of `"0.1 mol/L NaOH"` is 13. But after adding 80 mL of solution, you will have 30 mL of `"NaOH"` in 130 mL of solution. The `"pH"` will be less, say 12. Plot a point at (`80, 12`).

Just past the equivalence point

Just as the `"pH"` started rising rapidly 5 mL before the equivalence point, it will start levelling off at 5 mL past the equivalence point. Plot a point at (`55, 11`).

Put them all together

Join all the points by a smooth line. It should look something the plot above.

Remember, this is just a sketch. If you have to plot a titration curve, you will have to calculate the points.