Buffer Question?

Ka for CH₃COOH = 1.8 × 10⁻⁵

Sodium acetate is added to the acetic acid solution to prepare the buffer.

Henderson-Hasselbalch equation:
pH = pKa + log([CH₃COO⁻]/[CH₃COOH])
4.0 = -log(1.8 × 10⁻⁵) + log([CH₃COO⁻]/1.0)
log[CH₃COO⁻] = -0.74
[CH₃COO⁻] = 10⁻⁰˙⁷⁴ M = 0.18 M

Molar mass of CH₃COONa = 12×2 + 1×3 + 16×2 + 23 = 82 g/mol
Mass of CH₃COONa added = (0.18 mol/L) × (50/1000 L) × (82 g/mol) = 0.74 g

To make the buffer, dissolve 0.74 g of CH₃COONa into the 50 mL of 1.0 M acetic acid.

The other answer gets to the proper buffer by adding sodium acetate.

Suppose that the problem explicitly says to add NaOH. How much should be added. Here's the set up:

4.0 = -log(1.8 × 10⁻⁵) + log(x / 1.0 - x)

What you are doing is creating some CH3COONa from the CH3COOH already there. So, the acetic acid goes down by the same amount the sodium acetate goes up.

log(x / 1.0 - x) = -0.7447

(x / 1.0 - x) = 0.18

0.18 - 0.18x = x

1.18x = 0.18

x = 0.15254 M

(0.15254) (50/1000) (39.9969) = 0.305 g