Can someone explain this chemistry please?

Question

#81)

Calculate the pH of each of the following solutions:

A) 0.0155 M HBr

B) 1.28 x 10^-3 M KOH

C) 1.89 x 10^-3 M HNO3

D) 1.54 x 10^-4 sr(OH)2

Okay, so I got the first one, but I don't get the last three.

Please explain so I can understand in the future. thanks much :)

Answer 1

Basically, to find pH, you have to -log the molarity.

So, for A and C you would do:

A) -log(0.0155) = pH

C) -log(1.89x10^-3) = pH

For B and D, you do the same -log technique. But, because there is an OH group, you just found out the pOH. To find the pH, you must subtract by 14.

pH + pOH = 14 (LEARN THIS)

B) 14 - (-log(1.28 x 10^-3)) = pH

D) 14 - (-log(1.54 x 10^-4)) = pH

Hope this helps and good luck :)))