First, please spell chemistry right. I'm annoyed. Second, this is a buffer problem for the first addition. They throw you a curve with the second addition because it is NOT a second buffer problem. Here is the scoop.
Call acetic acid, CH3COOH, as HAc.
millimols HAc = mL x M = 400 x 0.1 = 40.
mmols NaOH (first time) = mL x M = 20
mmols NaOH (second time) = 20
Ka = 1.8E-5; pKa = -log Ka = 4.74
Henderson-Hasselbalch equation is
pH = pKa + log (base)/(acid) and since the volumes are the same for each separate problem we can use millimoles and not actual concentrations (at least for the first addition).
.......HAc + OH^- ==> Ac^- + H2O
I......40....0.........0......0
add.........20.................
C.....-20...-20.......20......
E......20....0........20......
Then use the HH equation, plug the E line into the HH equation and solve for pH. The answer is 4.74.
For the second addition of 20 mmols NaOH we do this and START with the end of the first one.
.......HAc + OH^- ==> Ac^- + H2O
I......20....0........20......
add..........20................
C.....-20...-20.......20......
E.......0....0........40
And here is the catch. Notice that the second addition of NaOH has neutralized ALL of the HAc and there is no acid left so this is no longer a buffer problem. What do we have in the solution? Just 40 millimols Ac^- (NaAc or sodium acetate) in 400 mL + 100 mL + 100 mL H2O so (Ac^-) = mmols/mL = 40/600 = 0.0667M NaAc. So the pH is determined by the hydrolysis of the NaAc, like this.
........Ac^- + HOH => HAc + OH^-
I.....0.0667...........0......0
C.......-x............x......x
E.....0.0667-x.........x......x
Kb for Ac^- = (Kw/Ka for HAc) = (x)(x)/(0.0667-x) and solve for x = (OH^-), then convert to pH. I didn't work this particular problem but most acetate hydrolysis problems come out to be in the pH range of 8.3 to 9. With a starting concn of 0.0667 sometimes you must use the quadratic equation (that is, 0.0667-x must be used and you can't ignore the -x) so make sure the quadratic is not required.
Post your work if you run into trouble with this and explain in detail what you don't understand about it.
To 400mL of a 0.1 M CH3COOH solution(Ka=1.8 x 10-5),two subsequent additions of 100mL of 0.2 M NaOH solution have been done.determine the solution pH after each addition.
Drbob,I don't understand this problem how to solve.Can you help me and give me the full calculation.
2 answers
A quadratic is not required and you can let 0.0667-x = 0.0667.
2 answers