Posted by millie on Wednesday, October 7, 2009 at 9:20pm.
I assume you are using pKa = 4.74 so that
pH = 4.74 + log 0.5/0.5 =
= 4.74 + log 1 = 4.74 + 0 = 4.74.
If we call acetic acid HAc and the acetate ion (the base) Ac, then
To make the pH 5.74 we calculate the B/A as follows:
5.74 = 4.74 + log B/A
You can go through the calculation but you should end up with (Ac)/(HAc) = 10
Then we write the equation
HAc + NaOH = NaAc + H2O
HAc = 0.5 to start
Ac = 0.5 to start.
We add x moles NaOH.
The HAc after the reaction will be 0.5 - x and the Ac will be 0.5+x
Just plug those values into the
(Ac)/(HAc) = 10 and solve for x.
I always like to put that value in and resolve the equation to see if I get 5.74.
Post your work if you get stuck.
thanks for your help, it worked out :)
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