A) A beaker with 1.80×102 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 5.50 mL of a 0.340 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.
B)You need to produce a buffer solution that has a pH of 5.37. You already have a solution that contains 10. mmol (millimoles) of acetic acid. How many millimoles of acetate (the conjugate base of acetic acid) will you need to add to this solution? The pKa of acetic acid is 4.74.
(A) Total concentration = 0.100 M
[acetic acid] + [acetate] = 0.100 M ...(1)
According to Henderson - Hasselbalch equation,
pH = pKa + log([conjugate base] / [weak acid])
pH = pKa + log([acetate] / [acetic acid])
5.000 = 4.740 + log([acetate] / [acetic acid])
log([acetate] / [acetic acid]) = 5.000 - 4.740
log([acetate] / [acetic acid]) = 0.260
[acetate] / [acetic acid] = 100.26
[acetate] / [acetic acid] = 1.82 ...(2)
Solving equations (1) and (2),
[acetic acid] = 0.035465 M
[acetate] = 0.064535 M
moles acetic acid = (concentration acetic acid) * (volume of buffer)
moles acetic acid = (0.035465 M) * (180 mL)
moles acetic acid = 6.383656 mmol
Similarly, moles acetate = 11.616344 mmol
moles HCl added = (concentration HCl) * (volume HCl)
moles HCl added = (0.340 M) * (5.50 mL)
moles HCl added = 1.87 mmol
new moles acetic acid = (initial moles acetic acid) + (moles HCl)
new moles acetic acid = (6.383656 mmol) + (1.87 mmol)
new moles acetic acid = 8.253656 mmol
new moles acetate = (initial moles acetate) - (moles HCl)
new moles acetate = (11.616344 mmol) - (1.87 mmol)
new moles acetate = 9.746344 mmol
Again using Henderson - Hasselbalch equation,
pH = pKa + log([acetate] / [acetic acid])
pH = 4.740 + log(9.746344 mmol / 8.253656 mmol)
pH = 4.812
pH change = (final pH) - (initial pH)
pH change = (4.812) - (5.000)
pH change = -0.188
(B) initial moles acetic acid = 10. mmol
According to Henderson - Hasselbalch equation,
pH = pKa + log([acetate] / [acetic acid])
5.37 = 4.74 + log(moles acetate / 10. mmol)
log(moles acetate / 10. mmol) = 5.37 - 4.74
log(moles acetate / 10. mmol) = 0.63
moles acetate / 10. mmol = 100.63
moles acetate / 10. mmol = 4.27
moles acetate = 10. mmol * 4.27
moles acetate = 42.7 mmol
millimoles of acetate to be added = 43 mmol
A) A beaker with 1.80×102 mL of an acetic acid buffer with a pH of 5.000...
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