how many mL of .100 M of acetic acid and .100 M of NaOH, would be need to make the volume 1.0 liters and have a pH7.00 if Ka=1.8x10^-5
please write out the steps or eq for me im gettin stuck
chem - DrBob222, Thursday, April 8, 2010 at 10:37pm
This is not a good question because the pH of 7.00 is too far from the pKa value to work well as a buffer. However, here is the problem.
pH = pKa + log (base)/(acid)
7.00 = 4.74 + log (B/A)
Solve for (base/(acid). I obtained 181.97 but you need to confirm that. Here base is referring to acetate ion (not NaOH) and acid is referring to acetic acid. I call those HAc for acetic acid and Ac^- for acetate (the base).
Then you set up another equation.
moles base = 181.97*moles acid
For moles base substitute mLbase x 0.1 M. On the right side, substitute for moles acid (1000 - mLbase)*0.1 M
Solve for mLbase
I get something like 994.53 for base and 1000-994.53 = 5.47 mL acid. You need to confirm both numbers AND you may need to carry out to more places that you are allowed under significant figure rules BECAUSE it is such a large number for one and a small number for the other one.
What this means is that you need to add that many mL of 0.1 M NaOH to produce that much Ac^- and since it is neutralizing HAc, the HAc is being used up at the same time. To check if this is correct,
(Ac^-) formed = 994.53 mL x 0.1M/1000 mL = 0.099453 M.
(HAc) remaining = 5.465 mL x 0.1 M/1000 = 0.0005465
pH = 4.74 + log (0.099453/0.0005465) =
pH = 4.74 + 2.26 = 7.00. I reiterate this is a poor problem AND that I have used more significant figures than allowed. With such a disparity in numbers, that is necessary to make the numbers come out right.
chem - phil, Friday, April 9, 2010 at 3:33pm
I don't think this would be correct because The solution you are proposing would have a pH over 12.