Whats the pH after 100.00 ml M Na OH is added to 1.00 L of buffer HA=1.45 M, A=0.125 M pKa 4.98.initial Ph3.92

Question

Whats the pH after 100.00 ml M Na OH is added to 1.00 L of buffer HA=1.45 M, A=0.125 M pKa 4.98.initial Ph3.92

Answer 1

Sorry, I can't understand the way you worded the question... If M is moles, it would help if you wrote that out, but it's still a little confusing. Is the buffer HA? The second line just doesn't make sense.

Answer 2

pH = 4.20

No solution will be provided.

Just remember that

[H3O+] = Ka * (acid/base)

and

pH = -log([H3O+])

Your reaction after adding NaOH would be

HA + OH- <--> A- + H2O

Also try using an ICE table with changes in moles and not molarity.

[Answer: see above]

NOTE: THIS IS A REPEATED QUESTION.