$100 mL$ of $0.20 M$ $CH X_{3} COONa$ ( $p K_{a} = 4.74$ for acetic acid) has pH $= 9.02$ . Which of the following changes give the stated pH? (Assume additive volumes; for the excess-base case ignore the hydrolysis of acetate.)

Answer: A, B, C

Water:

pH = 7 + \frac{1}{2} (4.74 + lo g 0.10) = 8.87

. NaOH:

[OH X^{-}] = 0.10 M

gives pH

13.00

0.10 M

HCl:

10 mmol

HCl converts half of the

20 mmol

acetate to acetic acid, a 1:1 buffer,

pH = p K_{a} = 4.74

. With

0.20 M

HCl all the acetate becomes

0.10 M

acetic acid, a weak acid with

pH = \frac{1}{2} (p K_{a} + 1) = 2.87

, not

p K_{a}