[MrTom]

Hi all, first post here.

This seems pretty easy, but I came across this question, and wondered how I would go about answering it:

Calculate the pH of the resulting solutions when:

1) 10.0cm3 of 1.0M HCl are added to 990cm3 of water
2) 20.0cm3 of 1.0M NaOH are added to 980cm3 of water.

Thanks alot in advance.

MrTom

[Borek]

1^st - out of the pH definition.

2^nd - out of the pOH definition (assuming you know what pH+pOH is).

Please read forum rules.

[MrTom]

Ooh, ofcourse, I feel abit dumb now.
I'm quite new to the topic and I was getting abit 'led astray' by the water in the question, but I got it now;

1) Work out the number of moles of HCl; 0.01dm3 x 1.0M = 0.01 moles HCl, therefore, 0.01 moles H⁺
pH = -log100.01 = 2

2) Again, work out moles of NaOH (0.02 x 10.M =) 0.02 moles.
pKa = -log₁₀0.02 = 1.7
pKa + pKb = 14, so, pKb = 14 - 1.7 = 12.3