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Here’s how you sketch the titration curve.
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 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.
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