Chemistry Help Please!!!?

Question

How do you:

#1: Calculate the pH of 0.075 M KOH.

I tried -log[7.5x10^-2] but that didn't work...

#2: The pH of a solution of hydrochloric acid is 2.80. What is the molarity of the acid?

Thanks a lot for those who help = )

Answer 1

Problem 1 involves a strog base and Problem 2 involves a strong acid.  This makes solving them really easy.

#1

Since KOH is a strong base, then the [OH-] will be 0.075.

pOH = - log [OH-] = 1.12

pH + pOH = 14

pH = 14 - 1.12 = 12.88

#3

Since HCl is a strong acid, the molarity of the H+ will be the same as the molarity of the acid.

pH = - log [H+]

- pH = log [H+]

-2.80 = log [H+]

antilog (-2,80) = [H+]

1.56 x 10^-3 = [H+] = [HC]

Answer 2

Ok so, let's see.

1) You were close, but when you do what you did, you are finding the pOH since KOH is a strong base. Since pOH + pH = pKw = 14, so all you need to do is:

14 - (-log(.075)) = 14 - 1.12 = 12.88

This is a high pH, which makes sense since (again) KOH is a strong base.

2) HCl is a strong acid and, as such, it dissociates completely in solution, forming H+ and Cl-.

pH = -log[H+]

2.80=-log[H+]

-2.80=log[H+]

10^(-2.80) = [H+] = 0.00159 M

The [H+] = the concentration of the acid.

I hope that helps :)