What is the pH when enough 0.10 M Base NaOh (in mL) is added to neutralize 8 mL of 0.10 M Acid HCl? Acid HC2H3O2?
I got the pH of HCl to be 7 because I figured that the same molarity of base and acid are added together to neutralize the pH, but I keep messing up for HC2H3O2. I don't know if there is a different method for this, or if my method is wrong.
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)]
does pH = pKa? because I remember my professor mentioning something about that in class.
sorry...that was a dumb question
would the pKa = pH in this case? because the ml and concentrations are equal in each scenario? If you plug it in the formula, I think your logs would equal 0 because it would be log(1) = 0. My asnwer is not coming out right though.
pKa is the pKa of the acid, in this case it is acetic acid.
pKa = -log Ka
pKa = - log(1.8 x 10^-5)
pKa = ??. I think it's about 4.7 something. The logs aren't equal. As I remember you have 0.00019 moles of the base and 0.00061 moles of the acid.