Question

hello, I can't seem to solve this question. i tried to look up someone else who may have asked this question. I found this girl named Sara,but her answer didn't really help me much because it did not answer the question.

What is the pH when enough 0.10 M NaOH Base (in mL) is added to neutralize 8 mL of 0.10 M HC2H3O2 Acid? HCl Acid?

I already solved for the pH of HCl and HC2H3O2 for the previous questions, but this question puzzles me because they do not really give you that much information to solve the question. What gets plugged into the pH = pKa + log(base/acid) equation? We do not have a value for pKa or the log(base/acid).
This is what was sent to Sara:
"When a strong acid is added (exactly neutralized) to a strong base, the salt produced is neutral (neither cation nor anion is hydrolyzed) and the pH = 8.

When a weak acid and a salt of the weak acid are present in solution, you hve a buffered solution and you must use the Henderson-Hasselbalch equation to solve for the pH.
pH = pKa + log [(base)/(acid)]"

Answer 1

Wouldn't the pH be 7 for the strong acid, not 8? I'm confused....

Answer 2

Yes, the pH is 7. I made that post and I hit the wrong key. Of course 7.0 is the correct answer and not 8 for NaCl.
For the other part, if you exactly neutralize NaOH and acetic acid, you will have at the equivalence point a solution of sodium acetate, the salt of a weak acid and a strong base.The acetate ion is hydrolyzed to obtain
C2H3O2^- + HOH ==> HC2H3O2 + OH^-

Set up an ICE chart and plug into the following.
Kb = (Kw/Ka) = (HC2H3O2)(OH^-)/(C2H3O2^-).
So X = (HC2H3O2) = (OH^-) and you know Kw and Ka. Concn of the starting salt, C2H3O2^-, will be 0.05 M (that's 1/2 x 0.1 M = 0.05 M).
Solve for OH, convert to pOH, then to pH.

Answer 3

When you set up the ICE chart, are you supposed to bring the 0.05 to the product side to give you an equation of x^2 / x - 0.05?