Answer to Question #292482 in General Chemistry for kate

Question

Calculate the pH of a solution which contains 2.0 x 10-3 M sodium acetate and 3.3 x 10-2 M acetic acid in a liter solution. The pKa is 4.74.

Answer 1

I'll assume that the remainder fo the problem, which is cut off, is "assume the volume doesn't change."

Also, to answer this, we need to know the Ka or pKa for acetic acid. The normal value given for the pKa is 4.75. Note, this is the same as the desired pH. This becomes important.

This is a BUFFER problem. Most buffer problems can best be addressed using the Henderson Hasselbalch equation:

pH = pKa + log [salt]/[acid] MEMORIZE THIS.

We know the pH (4.75)

We know the pKa (4.75). 

We know the [acid] = 0.20 M

We can find [salt] ,i.e [sodium acetate]

When the pH = pKa as in this case, the ratio of [salt]/[acid] =1 because log 1 = 0 

So, [sodium acetate] = 0.20 M. This can be seen as follows:

4.75 = 4.75 + log [sodium acetate]/[0.20]

0 = log x/0.2

x = 0.2 M

So, to now calculate the "amount" in grams, you would convert 0.20 M to grams using the volume of 200 ml (0.2 L)

0.20 mole/L x 0.2 L = 0.04 moles moles sodium acetate

0.04 moles sodium acetate x 82 g/mole = 3.3 g (to 2 significant figures)