Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.

QUESTION

How do you draw a general titration curve for titrating a weak acid with a strong base (let ph of equivalent point be 9.00)?

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.