- #1
nobb
- 33
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Hi.
I am having trouble with acid-base titration excercises. 0.02 L butanoic acid @ 0.1 M is titrated with 0.01 L NaOH solution @ 0.1 M. The Ka of butanoic acid is 1.54X10^-5. Find the pH. The answer is 4.81. I know how to find the pH when no NaOH is added, but I really don't see how the Ka fits in when adding a base. An explanation would really be appreciated. Thanks.
I am having trouble with acid-base titration excercises. 0.02 L butanoic acid @ 0.1 M is titrated with 0.01 L NaOH solution @ 0.1 M. The Ka of butanoic acid is 1.54X10^-5. Find the pH. The answer is 4.81. I know how to find the pH when no NaOH is added, but I really don't see how the Ka fits in when adding a base. An explanation would really be appreciated. Thanks.