A beaker with 1.90×102 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 8.40 mL of a 0.410 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.
pH of buffer
pH = pKa + log [salt/[acid]
5.000 = 4.740 + log [salt/[Acid]
Or, log [salt /acid] = 5.000 - 4.740 = 0.26.
Or, [salt/acid] = 100.26 = 1.819.
Given, [ salt ] + [ acid] = 0.100 M
Or, 1.819 [acid] + [acid] = 0.100
Or, 2.819 [acid] = 0.100
Or, [acid] = 0.100/2.819 = 0.0354 M
Then [salt] = (0.100 - 0.0354) = 0.0646 M.
milmoles of acetic acid = 0.0354* 1.90*102 = 6.726
milimoles of salt = 0.0646*1.90*102 = 12.274
milimoles of HCl added = 0.410*8.40 = 3.444
The reaction when HCl is added
CH3COONa + HCl CH3COOH + H2O
now ice table is
CH3COONa | HCl | CH3COOH | |
Initial milimoles | 12.274 | 3.444 | 6.726 |
Change | -3.444 | -3.444 | +3.444 |
equilibrium milimoles | 8.827 | 0 | 10.17 |
pH = pKa + log (8.827/10.17)
Or, pH = 4.74 + log ( 0.8679)
Or, pH = ( 4.74 - 0.0615) = 4.678
So, pH in change by (5.000 - 4.678) = 0.322
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