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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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