In: Chemistry
BUFFER PROBLEMS
3. Start with either the conjugate acid or base and add a strong base or strong acid (conjugate acid plus strong base forms the conjugate base and water; conjugate base plus strong acid forms the conjugate acid): What is the pH of 0.5 L of a 0.1 M acetic acid solution to which 0.73 g of NaOH are added? The pKa of acetic acid is 4.76.
moles of CH3COOH = molarity x volume
= 0.1 x 0.5
= 0.05
moles of NaOH = mass / molar mass
= 0.73 / 40
= 0.01825
CH3COOH + NaOH ---------------------> CH3COONa + H2O
0.05 0.01825 0 0 -----------------> initial
-0.001825 - 0.01825 + 0.01825 + 0.01825 ---------------> change
0.03175 0 0.01825 0.01825 --------------->equilibrium
pH = pKa + log [CH3COONa / CH3COOH]
pH = 4.76 + log ( 0.01825 / 0.03175)
pH = 4.52