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Oksana_7056
#1 Posted : Friday, November 19, 2021 7:15:27 PM
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Lesson 7: Question 8 in classroom companion (pg 360). Why was B also the correct answer?

If I remember correctly, Katrina's explanation was something similar to the following:

You can make a buffer indirectly by mixing lots of weak base and a small amount of strong acid (or lots of weak acid and small amount of strong base). This involved two steps. First, neutralization to H2O and acidic salt. Second step would then be remaining weak base (CH3COONa) + H2O = conj. A (CH3COOH) + hydroxide ion.

What confuses me is the 1st step above, specifically the neutralization part. I don't see how water and acidic salt are produced in step 1.

I thought it was just: CH3COONa + HCl = CH3COOH + NaCl, where excess/remaining CH3COONa and its conjugate acid CH3COOH form the buffer, and NaCl is just neutral salt.

Please correct me if I am wrong. Thanks.
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INSTR_Molly_129
#2 Posted : Tuesday, November 23, 2021 12:45:49 AM
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Hi Oksana!

Great question. In general, you can make a buffer whenever there is presence of weak acid and its conjugate base, or vice versa.

In the case of "B", CH3COONa is actually, CH3COO- and Na+. Remember that salts are spectator ions in buffer solutions, and not necessary to consider. Therefore, when you mix in HCl, you get exactly what you had wrote down: CH3COONa + HCl = CH3COOH + NaCl

Now think of it the way I just mentioned, you really have: CH3COO- + Na+ + HCl = CH3COOH + NaCl

Since you have both the weak base and its conjugate acid, you have a buffer in the making!

Hope this helps, and happy studying!

Molly
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