Answer to Question #103247 in General Chemistry for Jonathan

Question

The flask contains 25 mL of an unknown diprotic acid aqueous solution that reacts in a 1:2 stochiometric ratio with NaOH. Titrate the solution with NaOH to determine the concentration of the acid.

Answer 1

First, we need to know the concentration of NaOH. The problem can't be solved without this information. Assume it is 0.1 mol/l.

Second, we add phenolphthalein to the acid solution. It will not change the color of the acid and the water, but it will make the solution pink if there are any free molecules of NaOH.

Then we titrate the solution with NaOH: we use a burette to add known volumes (drops) of the base unless the acid aqueous solution becomes pink. Assume that each drop of NaOH solution has a volume of 0.05 ml.

As the acid solution becomes pink, it means that all molecules of the acid reacted with all molecules of NaOH and the excess of NaOH reacted with phenolphthalein. Assume that the initial volume of the solution in the burette was 30 ml, the final volume (when the solution in conical flask turned pink) in the burette was 23 ml.

Hence, the volume of NaOH is 7 ml. It means that there were

$$n_\text{NaOH}=\frac{V_\text{NaOH}}{c_\text{NaOH}}=\frac{0.007\text{ l}}{0.1\text{ mol/l}}=0.07\text{ mol}$$

of the pure base.

Write the supposed equation:

H₂SO₄ + 2NaOH => Na₂SO₄ + 2H₂O

It means that each molecule of the acid required 2 molecules of the base to react. Therefore, it means that the amount of the acid involved in the reaction was 0.035 mol, two times less.

The volume of the acid was 25 ml. Therefore, the concentration is

$$c_{\text{H}_2\text{SO}_4}=\frac{n_{\text{H}_2\text{SO}_4}}{V_{\text{H}_2\text{SO}_4}}=\frac{0.035\text{ mol}}{25\cdot10^{-3}\text{l}}=1.4\text{ mol/l}.$$

Of course, this is approximate result due to the procedure of titration.