Asked by millie
a buffer composed of 0.50mol acetic acid and 0.50mol sodium acetate is diluted to a volume of 1.0L. The pH of the buffer is 4.74. How many moles of NaOH must be added to the buffer solution to increase its pH to 5.74?
Answers
Answered by
DrBob222
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.
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.
Answered by
millie
thanks for your help, it worked out :)