[Bob Sacamano]

1.5 mmol of NaOH is added to a 1L, 10mM solution of a weak acid with pKa = 0.95. What is the pH of the resulting solution?

I don't know what to do.

[rabolisk]

Heres a hint. HA (acid) + B (base) = salt + H2O.

[vmelkon]

Step 1 : how many molecules of NaOH? The answer is already given : 1.5mmol. But I prefer to write that as 0.0015 mol.
Step 2 : How many molecules of the acid do you have? This is something you must calculate.
I prefer to write 10 mM as 0.0010 M. We can also write it as 0.0010 mol/L.

So 1 L * 0.0010 mol/L = 0.0010 mol of acid.

Step 3 : What is the chemical reaction? This is not given to us.
Is it NaOH + H-something = salt + H20?
Is it 2 NaOH + H2-something = salt + 2H20?
Is it 6 NaOH + 2 H3-something = salt + 6H20?
Is it 4 NaOH + H4-something = salt + 4H20?

If it is the first, then we have more base than acid. Excess is 0.0005 mol in 1 L.
Can you compute the pH?

What if it is the second?

[Borek]

I prefer to write 10 mM as 0.0010 M.

10 mM is OK with me, but if anything, I prefer to write it as 0.010M.

Please read forum rules. You are on the verge of giving too much help.

[vmelkon]

Sorry. (about the giving too much help part). I'll cut back.