You are given 60 mL of 0.50 M acetic acid/ acetate buffer to test. The starting

Question

You are given 60 mL of 0.50 M acetic acid/ acetate buffer to test. The starting composition of the two major species are:

Concentration of CH3COOH: 0.300 M

Concentration of CH3COO-: 0.200 M

A. Use the henderson hasselbach equation to estimate the initial pH of the buffer.

B. You add 1.0 mL of 1.00 M HCl to the buffer. Calculate the molarity of H3O+ added as HCl, and the final molarities of acetic acid and acetate ion at equilibrium. What is the new value of the pH?

C. Now take a fresh 60 mL of the buffer and add 1.0 mL of 1.00 M NaOH. Using steps similar to those above calculate the new pH of the solution.

Please show your work. Thanks

Explanation / Answer

a)

pH = pKb + log(acetate/Acid)

pH = 4.75 + log(0.2/0.3) = 4.5739

b)

mmol of HCl = 1*1 = 1 mmol

mmol of acetate = MV - 1 = 60*0.2 - 1 = 11

mmol oa cetic acid = MV +1 = 60*0.3 + 1 = 19

then

pH = 4.75 + log(11/19) = 4.5126

c)

for the same amount of base:

mmol of HCl = 1*1 = 1 mmol

mmol of acetate = MV + 1 = 60*0.2 + 1 = 13

mmol oa cetic acid = MV - 1 = 60*0.3- 1 = 17

then

pH = 4.75 + log(13/17) = 4.63349

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