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.
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