Write reaction equations to explain how your acetic acid-acetate buffer reacts w

Question

Write reaction equations to explain how your acetic acid-acetate buffer reacts with an acid and reacts with a base. Buffer capacity has a rather loose definition, yet it is an important property of buffers. A commonly seen definition of buffer capacity is: "The amount of H^+ or OH^- that can be neutralized before the pH changes to a significant degree." Use your data to determine the buffer capacity of Buffer A and Buffer B. Say, for example, that you had prepared a Buffer C, in which you mixed 8.203 g of sodium acetate, NaC_2H_3O_2, with 100.0 mL of 1.0 M acetic acid. What would be the initial pH of Buffer C? If you add 5.0 mL of 0.5 M NaOH solution to 20.0 mL each of Buffer B and Buffer C, which buffer's pH would change less? Explain.

Explanation / Answer

1) the equation will be

a) reaction with acid

HCl + CH3COONa --> CH3COOH + NaCl

b) reaction with base

NaOH + CH3COOH --> CH3COONa + H2O

2) Buffer capacity = dn / dpH

dn = number of equivalents of acid/ base added

dpH = change in pH

buffer capacity = (0.05 / 2) = 0.025

3) the pH of buffer is obtained from Henersen Hassalbalch equation as

pH = pKa + log [salt] / [acid]

moles of salt added = Mass of salt / molecular weight of salt = 8.203 / 82.03 = 0.1 moles

Moles of acid added = Molarity X volume in litres = 1 X 0.1 = 0.1

pH = pKa + log [0.1] / [0.1]

pKa = 4.74

pH = 4.74

b) if we add 5mL of 0.5M NaOH then the moles of base added = Molarity X volume

                   = 0.5 X 5 millimoles = 2.5millimoles

For Buffer C

The NaOH will react with 2.5millimoles of acid tof form 2.5millimoles of salt

final moles of acid =100-2.5 = 97.5 millimoles

Final moles of salt = 100+ 2.5 = 102.5 millmoles

pH = 4.74 + log [102.5 / 97.5 ] = 4.76

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