First, the question I'm given is: What is the pH of a 1.0 L solution containing 0.25M acetic acid and 0.75M sodium acetate ( Ka for acetic acid= 1.8x 10-5)
So I took the -log(1.8x10-5)= 4.74+log(.75/.25)= 5.22 pH
But then I'm asked if .050 mol NaOH is added to the above solution, what is the new pH? (assume no change in volume) How would the new pH be calculated?
acetic acid = HAc
sodium acetate = NaAc
millimols HAc = 1000 mL x 0.25 = 250.
mmols NaAc = 1000 mL x 0.75 = 750.
0.05 mol NaOH = 50 mmols.
.......HAc + OH^- ==> Ac^- + H2O
I......250....0.......750.....0
added........50.................
C......-50...-50......+50....+50
E......200....0.......700........
(HAc) = 200/1000 = ?
(Ac^-) = 700/1000 = ?
Note that the denominator value of 1000 cancels.
pH = 4.74 + log(700/200) = ?