strong acid to buffer A beaker with 145 m L of an acetic acid buffer with a pH o

Question

strong acid to buffer A beaker with 145mL 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 7.90mL of a 0.360MHCl 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. strong acid to buffer A beaker with 145mL 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 7.90mL of a 0.360MHCl 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 change in pH = -0.35 (minus 0.35)


Let acetic acid be represented by HA: HA <=> H+ + A-

Initial pH = pKa + log ([A-]/[HA])

5.00 = 4.76 + log ([A-]/[HA])

[A-]/[HA] = 1.738 => [A-] = 1.738[HA]


Total molarity = [HA] + [A-] = 0.100 M

[HA] + 1.738[HA] = 2.738[HA] = 0.100 => [HA] = 0.03652 M

Moles of [HA] = 145/1000 x 0.03652 = 0.0052954 mol


[A-] = 1.738[HA] = 1.738 x 0.03652 = 0.06348 M

Moles of [A-] = 145/1000 x 0.06348 = 0.0092046 mol


After addition of HCl: HCl + A- => HA + Cl-

Moles of HCl added = 7.90/1000 x 0.360 = 0.002844 mol

Moles of HA = 0.0052954 + 0.002844 = 0.0081394 mol

Moles of A- = 0.0092046 - 0.002844 = 0.0063606 mol


Final pH = pKa + log([A-]/[HA])

= pKa + log(moles of A-/moles of HA)

= 4.76 + log(0.0063606/0.0081394) = 4.65


Change in pH = final pH - initial pH

= 4.65 - 5.00 = -0.35

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