Question
In: Chemistry
A beaker with 1.70×102 mL of an acetic acid buffer with a pH of 5.000 is...
A beaker with 1.70×102 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 5.70 mL of a 0.430 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.
Solutions
Expert Solution
pH = 5.00
pH = pKa + log [salt / acid]
5.00 = 4.74 + log[conjugate base / acid]
[conjugate base / acid] = 1.819
milli moles of [conjugate base + acid] = 1.70×102 x 0.1 = 17
[conjugate base / acid] = 1.819
[conjugate base + acid] = 17
1.819 acid + acid = 17
acid milli moles = 6.03
conjugate base millimoles = 10.97
millimoles of HCl (C) = 0.430 x 5.70 = 2.451
on addition of ’ C’ millimoles of acid to acidic buffer salt millmoles decreases and acid moles increases
so
pH = pKa + log [Salt –C/acid + C]
= 4.74 + log [10.97 - 2.451 / 6.03 + 2.451]
= 4.742
pH change = final pH - initial pH
= 4.742 - 5.000
= - 0.258
pH change = - 0.258
queen_honey_blossom answered 3 months agoRelated Solutions
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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 7.70 mL of
a 0.430 M HCl solution to the beaker. How much will the pH change?
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