A beaker with 155 m L of an acetic acid buffer with a pH of 5.00 is sitting on a
ID: 782409 • Letter: A
Question
A beaker with 155mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 8.40mL of a 0.460MHCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760. Express your answer numerically to two decimal places. Use a minus (-) sign if the pH has decreased. A beaker with 155mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 8.40mL of a 0.460MHCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.760. Express your answer numerically to two decimal places. Use a minus (-) sign if the pH has decreased.Explanation / Answer
The answer is: pH change = -0.77. See solution below:
Let acetic acid be represented by HA
HA <=> H+ + A-
Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
5.00 = 4.76 + log ([A-]/[HA])
[A-]/[HA] = 1.738 => [A-] = 1.738[HA]
Total buffer concentration = [HA] + [A-] = 0.100 M
[HA] + 1.738[HA] = 0.100 M
[HA] = 0.100/2.738 = 0.03652 M
Moles of HA = volume x concetration of HA
= 0.155 x 0.03652 = 0.0056606mol
[A-] = 0.100 - 0.03652 = 0.06348 M
Moles of A- = volume x concetration of A-
= 0.105 x 0.06348 = 0.0066654 mol
After addition of HCl:
HCl + A- => HA + Cl-
Moles of HCl added = volume x concentration of HCl
= 8.4/1000 x 0.46= 0.003864 mol
Moles of HA = initial moles of HA + moles of HCl added
= 0.0056606 +0.003864 = 0.0095246 mol
Moles of A- = initial moles of A- - moles of HCl added
= 0.0066654 -0.003864= 0.0028014 mol
pH = pKa + log([A-]/[HA])
= pKa + log(moles of A-/moles of HA) since volume is the same for both
= 4.76 + log(0.0028014 /0.0095246 )
=4.22852837844
Thus change in pH = final pH - initial pH
4.228 - 5.00 = -0.77147162156
The pH decreases by 0.77 units
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