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)]"

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

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.

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?

To solve this question, you need to use the Henderson-Hasselbalch equation, which is pH = pKa + log(base/acid). However, before we can use this equation, we need to determine the values for pKa, base, and acid.

In this case, we are given that 0.10 M NaOH is added to neutralize 8 mL of 0.10 M HC2H3O2 acid. This means that the HC2H3O2 acid is reacting with the NaOH base in a 1:1 ratio.

To find the base, we need to calculate the number of moles of NaOH added. Since 0.10 M NaOH means there are 0.10 moles of NaOH in 1 liter, we can convert this to moles by multiplying 0.10 M by the volume in liters. 8 mL is equivalent to 0.008 L, so the number of moles of NaOH added is 0.10 M x 0.008 L = 0.0008 moles.

Next, we need to find the acid. The acid is HC2H3O2, and we are given the concentration is 0.10 M and the volume is 8 mL. Again, we can convert the volume to liters by dividing by 1000. So the number of moles of HC2H3O2 is 0.10 M x (8 mL / 1000) L = 0.0008 moles.

Now that we know the base and acid, we can substitute these values into the Henderson-Hasselbalch equation: pH = pKa + log(base/acid). However, you mentioned that the question did not provide a value for pKa. Without knowing the pKa for HC2H3O2, it is not possible to calculate the pH using the Henderson-Hasselbalch equation.

If the question did not provide a pKa value, then it is likely that we are not expected to use the Henderson-Hasselbalch equation to solve for the pH. In such cases, it is important to carefully read the question and look for any other information that may be provided. If the information given is insufficient to solve the problem, it is possible that there may be errors or missing information, and it would be best to clarify with your instructor or the source of the question.