Calculate pH of a solution containing .1M HCl and .1M HOCl?

Question

please include some explanation of how you solved the problem, but you don't have to go overboard with it

answer is 1

Answer 1

HClis a strong acid and fully dissociated. HOCl is a very weak acid; dissociation negligible especially in the presence of H+(aq) from the HCl

Take it from there!

Answer 2

This looks like a Gen Chem 2 question... I'll have to assume that you have a text that explains acid-base equilibria, strong and weak acids and Ka's.

HCl is a strong acid, so it will dissociate in solution completely. In a 0.1 M solution of HCl, the H3O+ concentration will be 0.1 M, so pH = -log(0.1) = 1.

HOCl is a weak acid (Ka = 3.58 x 10^-8), and only dissociates slightly in solution. The concentration of H3O+ in a 0.1 M solution of HOCl will be 5.98 x 10^-5 M, and the pH will be about 4.2*.

If you add together the H3O+ concentrations for the acids, you'd expect a value of 0.1000598 M. Because the concentration of H+ from the HCl is so large relative to that from the HOCl, the last part of the number can effectively be ignored. As near as makes no difference, the pH is still 1 in the mixture. In fact, the contribution from the HOCl will be even less, because the H3O+ from the HCl drives the HOCl dissociation equilibrium backwards (Le Chatelier's principle), and the actual contribution to the H3O+ concentration from the HOCl is miniscule.

* How I calculated that: if you're familiar with an ICE box, this explanation should be easy to understand. It's pretty clear from the question that you're probably expected to know how to use them for the course you're taking. I'll assume you know what I'm talking about (I have to - there's really no other way to explain the math).

Using the equation:

HOCl(aq) + H2O (l)--> H3O+ (aq) + ClO- (aq).

Initial concentrations before dissociation are 0.1 M HOCl, 0 M H3O+ and 0 M ClO- . After dissociation, the concentration of HOCl will have dropped by some amount x (concentration = 0.1-x), and the concentrations of H3O+ and ClO- will each have risen by the same amount.

Substituting into the equation for the equilibrium constant gives:

Ka = 3.58 x 10^-8 = x^2/(0.1-x)

Solving the equation can be simplified by assuming that x is small relative to 0.1, and so 0.1-x is, to a good approximation, 0.1.

That means 3.58 x 10^-8 = x^2/0.1, so x^2 = 3.58 x 10-9.
Therefore x = 5.98 x 10^-5 M. x is the H3O+ concentration, so pH = -log(5.98 x 10^-5) = 4.22.

Answer 3

7.
The acid and base cancel each other, leaving the value of the water.
Water is not H2O...it is HOH. Now look at your chemicals.