Olga weighed out 8.00g of imidazole and dissolved it in 100.0mL of water. She th
ID: 888102 • Letter: O
Question
Olga weighed out 8.00g of imidazole and dissolved it in 100.0mL of water. She then added 75.0 mL of 0.650M HCl to this solution and mixed them together. She then added enough water to give a total volume of 0.500L. Show your work and/or briefly explain your answers for a-d.
a) What is the pH and concentration of Olga's final buffer?
b) For an experiment, Okga takes 250.0mL of her buffer and mixes it with 5.00mL of 2.50M KOH. What is the pH and concentration of her buffer after this addition?
c) For an experiment, Olga takes 250.0mL of her buffer and mixes it with 5.00mL of 2.50M HNO3. What is the pH and concentration of her buffer after this addition?
d) Can Olga's initial buffer absorb acid or base better without showing a significant change in its pH? Explain your answer.
Explanation / Answer
Mass of imidazole = 8.0 gm
Molar mass of Imidazole = 68.077 g/mol
Moles of imidazole = 8.0 / 68.077
= 0.1175 moles
Molarity of HCl = 0.650 M
Volume of HCl added = 75 mL = 0.075 L
moles of HCl added = 0.650 * 0.075
= 0.04875 moles
(a). Moles of OH- ions in excess = 0.1175 - 0.04875
= 0.06875 moles
Total volume of solution = 100 + 75 = 175 mL
= 0.175 L
[OH-] = 0.06875 / 0.175
= 0.3928 M
pOH = - log [OH-]
= - log 0.3928
= 0.405
pH = 14 - pOH
= 14 - 0.405
pH = 13.595
(b). Volume of buffer taken = 250 mL = 0.250 L
Moles of OH- taken from buffer = 0.3928 * 0.250
= 0.0982
Moles of OH- ions added from KOH = 2.50 * 0.005
= 0.0125
Total moles of OH- ions = 0.0982 + 0.0125
= 0.1107
Total volume of buffer = 0.250 + 0.005
= 0.255 L
[OH-] = 0.1107 / 0.255
= 0.4341 M
pOH = - log 0.4341
= 0.362
pH = 14 - 0.362
pH = 13.638
(c). Moles of OH- taken from buffer = 0.3928 * 0.250
= 0.0982
Moles of HNO3 mixed = 2.50 * 0.005
= 0.0125
Moles of OH- ions consumed by acid = 0.0982 - 0.0125
= 0.0857
Total volume of buffer = 0.250 + 0.005
= 0.255 L
[OH-] = 0.0857 / 0.255
= 0.336 M
pOH = - log [OH-]
= - log 0.336
pOH = 0.473
pH = 14 - 0.473
pH = 13.527
(d). Olga's initial buffer can absorb base better without showing a significant change in its pH because of the excess mount to conjugate acid present in the buffer.
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