What is the pH of the final solution? ?

Question

A buffer solution is made by adding 75.0 g sodium acetate (NaC2H3O2) to 500.0 mL of a

0.64 M solution of acetic acid (HC2H3O2). What is the pH of the final solution? Assume no volume change.

Answer 1

Moles = mass/molar mass

Moles of NaC2H3O2 = 75/82=0.88 mol NaC2H3O2

Moraity = moles/volume

moles = molarity*volume(liter)

Moles of Acetic acid = 0.64M*(500/1000)L=0.32 mol

pKa of acetic acid = .

According to Henderson Hasselbalch Equation

pH = pka + log[salt]/[acid]

pH=4.75+log[ NaC2H3O2]/[C2H4O2]

pH=4.75+log(0.88/0.32)

pH=5.19

Answer 2

You need to know the molarity of the NaAc:

MV = mass / molar mass

(x) (0.5000 L) = 75.0 g / 82.0337 g/mol

x = 1.82852 M (keep some guard digits)

Now, use the Henderson-Hasselbalch equation:

pH = pKa + log [base / acid]

pH = 4.752 + log [1.82852 / 0.64]

pH = 4.752 + 0.456

pH = 5.208 (round off more as you see fit)