posted by UN on .
What is the pH of the solution created by combining 2.80 mL of the 0.10 M NaOH(aq) with 8.00 mL of the 0.10 M HCl(aq)? with 8.00 mL of the 0.10 M HC2H3O2(aq)?
mL of NaOH pH w/HCl pH w/HC2H3O2
2.80 ? ?
okay, so i tried m1v1=m2v2 and when i got m2 i would input it in -log(H+) to get the ph but it is wrong? the question asks for two reachants one with only HCl and then HC2H3O2 added as well.
Two questions. The point of these two problems is to show the difference between neutralizing a strong base with a strong acid versus neutralizing NaOH (a strong base) with a weak acid (HC2H3O2).
NaOH + HCl ==> NaCl + HOH
So you look to see what you have when mixing these two.
L x M = mols NaOH
L x M = mols HCl
Which is in excess? I think that is HCl. So HCl + NaCl is what you have in the end. Determine pH from that solution remembering that NaCl is not hydrolyzed AND that the strong acid is 100% ionized. pH = -log(HCl).
Second problem is slightly different.
NaOH + HC2H3O2 ==> NaC2H3O2 + HOH
L x M = mols NaOH
L x M = mols HC2H3O2
What do we have at the end? We have a salt of a weak acid (sodium acetate) + some of the weak acid that isn't neutralized. That creates a buffered solution and you can use the Henderson-Hasselbalch equation to solve it.
pH = pKa + log[(base)/(acid)]
If the problem doesn't give a pKa for acetic acid you can look it up in your text. It is very close to 4.74
Post your work if you get stuck. I probably won't check back until tomorrow. It's past my bed time.
okay, i got 2.8E-4 mols of NaOH
and 8e-4 MOLS of acetic acid, so do i just plug in mols of hcl into -log(hcl)?
In the first part as Dr BOB says the HCl is in excess (both HCl and NaOH have the same concentration but there is a greater volume of the HCl solution). Thus the excess moles of HCl we can get from the excess volume (8.00 - 2.80)ml = 5.20 ml. as the solutions are the same concentration.
So the number of moles of HCl is
5.20 x 10^-3 x 0.1 =5.20 x 10-4 moles
from which you can find the pH.
i inputed: -Log(5.20 x 10-4)
ph=3.28 but its incorrect?
You have omitted a step. Remember that (HCl) = mols/liter. You have mols from Dr Russ of 5.2 x 10^-4 mols. That is in a total volume of 2.80 mL + 8.00 mL = 10.80 mL or 0.0108 L.
Therefore, the final concentration of HCl = mols/L = 5.2 x 10^-4/0.0108 L = ?? then convert that to pH. Something like 1.3 or so. You need to do it and round appropriately.