Write reaction equations to explain how your acetic acid-acetate buffer reacts w
ID: 493267 • Letter: 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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