Question

How many moles of sodium acetate must be added to 500.0 mL of 0.250 M acetic acid solution to produce a solution with a pH of 4.94? (The pKa of acetic acid is 4.74)

A) 0.011 moles

B) 0.021 moles

C) 0.13 mol

D) 0.20 mol

E) 0.21 mol

Answer 1

(D)

Concentration of acetic acid = 0.250 M = 0.250 mol/L

Volume of acetic acid = 500.0 ml = 500 L / 1000 = 0.5 L

Number of moles of acetic acid = 0.5 L * 0.250 mol/L = 0.125 mol

Suppose moles of Sodium acetate added = x mole

From Henderson equation

pH = pKa + log { (sodium acetate/ acetic acid)}

4.94 = 4.74 + log {(sodium acetate / acetic acid)}

4.94 - 4.74 = log (x/0.125)

0.20 = log (x / 0.125)

x/ 0.125 = 10^0.20 = 1.585

x = 0.125 * 1.585 = 0.198 ~ 0.2

Number of moles of Sodium acetate added = x mole = 0.20 mole