Equalibrium and pH question :)?

Question

1. The OCl- ion acts as a base in water according to the equation

OCl-(aq) + H2O(l) HOCl(aq) + OH-(aq)

When two drops of 5.0 M NaOH are added to an equilibrium mixture of OCl in water at constant

temperature what is the effect on equilibrium?

2. Benzoic acid, C6H5COOH, molar mass 122 g mol–1, is a weak monoprotic acid.

The pH of a solution formed when 500 mg of benzoic acid is dissolved completely in water to form 200 mL

of solution is?

Answer 1

1. By adding NaOH into the mixture at equilibrium you disrupt the balance.  This is because NaOH dissociates into Na+ and OH-.  By increasing the concentration of OH-, by Le Chatelier's principle, you will then force the reaction to the left, thus you will get some OCl- and water forming.

2. First find out the mols of benzoic acid dissolved by dividing mass by molecular weight:

500 mg * (1g/1000mg) / 122g/mol = .0041mol BA

Now since it is monoprotic and completely dissolved we can assume:

C6H5COOH > C6H5COO- + H+

Therefore the mols of H= you get is equal to the mols of benoic acid.  Now divide this amount by the volume to get concentration of H+:

[H+] = .0041mol BA / 0.2 L = .0205 mol/L

Assuming this is an ideal mixture we can calculate the pH as follows:

pH = -log([H+]) = -log(0.0205) = 1.68

Hope this helped!