# Acid base titration

#### Maverick56

A 200.0 mL solution of 0.40 M CH3COOH(aq) and 0.40 M NaCH3COO(aq) has 20.0 mL of 1.0 M HCl(aq) added to it. What is the pH after the HCl has been added? (Ka = 1.8 x 10-5 for CH3COOH)

#### BerkReviewTeach

##### Company Rep & Bad Singer
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You start with 80 mmole each of HA and A-, which makes it a buffer with a pH = pK[sub]a[/sub] (4.74 in this case). To that you've added 20 mmoles of HCl, so after reaction you have a buffer with 60 mmoles A- and 100 mmoles HA (rich in conjugate acid). It makes the pH slightly less than the pK[sub]a[/sub], but not as much lower as log 2. So we know the pH is less than 4.74 but not as low as 4.44 (found from 4.74 - 0.30). You're looking for the answer choice in that range, but closer to 4.44 than 4.74.

What are the four answer choices that come with this MCAT practice question?

#### pjk77

How do you know that its "not as much lower as log 2?"

#### Next Step Tutor

##### MCAT Guru
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5+ Year Member
How do you know that its "not as much lower as log 2?"
He used the Henderson-Hasselbach equation.

Here's a picture with a full rundown:

http://i.imgur.com/sS3av3C.png

The key thing to recognize here is that you're starting with a buffer of a pH between 4.5 and 5 (you wouldn't be expected on the MCAT to be able to calculate this exactly) and that you're not adding that much acid - so you're still in the buffer region. Meaning the pH is still close to where you started.

#### pjk77

He used the Henderson-Hasselbach equation.

Here's a picture with a full rundown:

http://i.imgur.com/sS3av3C.png

The key thing to recognize here is that you're starting with a buffer of a pH between 4.5 and 5 (you wouldn't be expected on the MCAT to be able to calculate this exactly) and that you're not adding that much acid - so you're still in the buffer region. Meaning the pH is still close to where you started.

ahh ok-makes sense! Thanks!!