Calculate the pH of an aqueous solution of the following:

Question

2 x 10^-8 M KOH

So, for this I am getting the -log(2x10^-8) and I am getting 7.69 but the answer is suppose to be 7.1?

Could anyone show how they get to the correct answer? I think I am doing something wrong with bases because all of the acid questions I am getting the right pH but the bases are giving me a problem.

Thanks!

Answer 1

When the KOH concentration is that low, we MUST also consider the [OH^-] from the atuolysis of H₂O, which will be 1x10^-7 M (at 25ºC). So, 2x10^-8 M OH^- + 1x10^-7 M OH^- = 1.2x10^-7 M OH^-

pOH = -log 1.2x10^-7 = 6.9

pH = 14 - 6.9

pH = 7.1

Answer 2

Justin, the answer you quoted is also wrong.

Remember the pH = -log H+ions concentration. KOH is a strong base. When it ionises in solution it produces 2.0 x10^-8 moles of OH¯ ions. Another concept you are missing is that pH +pOH =14 .

If you have the OH ¯ ion concentration the –log of that gives you the pOH. To find the pH you will need to subtract that value from 14 to get the pOH.

I got 6.30 as the answer.–

-log 2.0 x10 ^-8 = 7.70 ( pOH).

pH =14-7.70 = 6.30.

Hopefully this helps for your future problems.