Adding a Strong Acid to a Buffer?

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2018-03-28 12:31 am

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A beaker with 155 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 4.30 mL of a 0.470 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740. Express your answer numerically to two decimal places. Use a minus ( − ) sign if the pH has decreased.

回答 (1)

wanszeto

2018-03-28 1:36 am

Assume that the buffer contains y M of the acid HA and (0.100 - y) M of the conjugate base A⁻.

HA(aq) ⇌ H⁺(aq) + A⁻(aq) …. pKa = 4.740
Henderson-Hasselbalch equation:
pH = pKa + log([A⁻]/[HA])
5.000 = 4.740 + log{(0.100 - y)/y}
log{(0.100 - y)/y} = 0.260
(0.100 - y)/y = 10⁰˙²⁶⁰
0.100 - y = 10⁰˙²⁶⁰y
(1 + 10⁰˙²⁶⁰)y = 0.100
y = 0.100 / (1 + 10⁰˙²⁶⁰)
y = 0.0355
[HA] = 0.0355 M
[A⁻] = (0.100 - 0.0355) M = 0.0645 M

After HCl is added, 1 mol of HCl reacts with 1 mole of A⁻ to give 1 mole of HA.
Final volume = (155 + 4.30) mL = 159.3 mL
[A⁻] = {0.0645 × (155/159.3) - 0.470 × (4.3/159.3)} M = 0.0501 M
[HA] = {0.0355 × (155/159.3) + 0.470 × (4.3/159.3)} M = 0.0472 M
Apply Henderson-Hasselbalch equation again:
pH = pKa + log([A⁻]/[HA])
pH = 4.740 + log(0.0501/0.0472)
pH = 4.766

Change in pH = 4.766 - 5.000 = -0.234 ≈ -0.23 (to 2 decimal places)

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