Question

Calculate the ph when 1.03 g of ch3coona (fw = 82.03 g/mol) is added to 36 ml of 0.500 m acetic acid, ch3cooh. ignore any changes in volume. the ka value for ch3cooh is 1.8 x 10-5.

Answer 1

moles of CH3COONa = mass / molar mass
when mass = 1.03 g and molar mass = 82.03 g/mol
So by substitution:
= 1.03 / 82.03 = 0.0126 moles

molarity of CH3COONa = moles / volume
= 0.0126 / 0.036 L
= 0.35 M
then we have to get Pka:
Pka = -㏒Ka
= -㏒1.8x10^-5
= 4.74
by using Henderson-Hasselbalch formula:

PH = Pka + ㏒[salt]/[acid]
PH = 4.74 + ㏒ 0.35 / 0.5
= 4.59