Question

Compute values for [H] [OH] and pH of the following aqueous solutions. Tabulate the answers for comparison. Spell out the reasoning in each calculation

a) 0.010M HCL
b)0.0010M NaOH
c) 0.010 N Ba(OH)2
(strong electrolytes assumed to be 100% dissociated.

Answer 1

Mostly the H+ is what is confusing me

Answer 2

(H^+) from HCl = (HCl) since HCl is 100% ionized. Then (H^+)(OH^-) = Kw = 1E-14
That allows you to calculate OH^-, the pH = -log(H^+).

For NaOH, it is 100% ionized; therefore, OH^- = (NaOH). Then solve for H^+ and pH.

Ba(OH)2 is a strong electrolyte; therefore, [Ba(OH)2] = 0.01 N, that is 0.01/2 = 0.005 M. (OH^-) is 2x[Ba(OH)2]. Then convert to H^+ and pH.

Answer 3

I still am lost on how to find OH- Kw being 1 x10^-14 could leave any possibility for H

Answer 4

For #1. 0.01 M HCl.
So H^+ = 0.01 M and since
(H^+)(OH^-) = 1E-14, then
(OH^-) = 1E-14/0.01 = 1E-12 M

For #2.
(NaOH) = 0.001 M so (OH^-) = 0.001 M
Then (H^+)(OH^-) = 1E-14
and (H^+) = 1E-14/0.001 = 1E-11 M

I'm confused by your question.
If you know (H^+) OR (OH^-) you immediately know the other BECAUSE
(H^+)(OH^-) ALWAYS equals 1E-14 (at least in water solution that is).