Question

An acetic acid-sodium acetate buffer can be prepared by adding sodium acetate to HCl(aq).

a) Write an equation for the reaction that occurs when sodium acetate is added to HCl(aq)

b) If 10.0g of CH₃COON is added to 275 mL of 0.225 M HCl(aq), what is the pH of the resulting buffer?

c) What is the pH of the solution in part b after the addition of 1.26g of solid NaOH?

Answer 1

a)

the reaction is given by

CH3COONa + HCl ---> CH3COOH + NaCl

b)

we know that

moles = mass / molar mass

so

moles of CH3COONa = 10 / 82 = 0.122

now

moles = molarity x volume (L)

so

moles of HCl = 0.225 x 275 x 10-3

moles of HCl = 0.061875

now

consider the reaction

CH3COONa + HCl ---> CH3COOH + NaCl

we can see that

moles of CH3COONa reacted = moles of HCl added = 0.061875

moles of CH3COOH formed = moles of HCl added = 0.061875

so

finally

moles of CH3COONa = 0.122 - 0.061875 = 0.060125

moles of CH3COOH = 0.061875

now

for buffers

pH = pKa + log [salt / acid ]

so

pH = pKa + log [CH3COONa / CH3COOH]

pH = 4.76 + log [ 0.060125 / 0.061875]

pH = 4.7475

so

the pH of the buffer is 4.7475

c)

moles of NaOH = 1.26 / 40 = 0.0315

now

the reaction is

CH3COOH + NaOH --> CH3COONa + H20

moles of CH3COOH reacted = moles of NaOH added = 0.0315

moles of CH3COONa formed = moles of NaOH added = 0.0315

finally

moles of CH3COOH = 0.061875 - 0.0315 = 0.030375

moles of CH3COONa = 0.060125 + 0.0315 = 0.091625

so

pH = pKa + log [CH3COONa / CH3COOH]

pH = 4.76 + log [ 0.091625 / 0.030375]

pH = 5.24

so

the pH of the solution after addition of NaOH is 5.24