What is the theoretical pH of a buffer solution that contains 50.0 ml of 1.0 HAc (aq) and 4.11 g of solid NaAc, and 25.0 ml of 1.0 NaOH (aq) thoroughly mixed together?
Number of moles of AcOH = 0.050L1 mole = 0.050
Number of moles of NaAc = 0.050L1 mole = 0.050 (82/1l;4.11g/50mL)
Ka = 1.7510-5
[HA] = [A-] = 1.0 mole0.050L = 20 moleL
According to Henderson-Hasselbach equation,
pH = pKa+log [A-] [HA] = 4.757+log20 = 4.757+1.301 = 6.058
Number of moles of base added = 0.025L1 mole0.050L = 0.5 moles
The added base will decrease the amount of AcOH and increase the amount of NaAc by 0.5 moles.
Therefore,
pH = pKa+log [A-] [HA] = 4.757+log[1.0][0.5] = 4.757+log2 = 4.757+0.3010 = 5.058
Change in pH, pH = 6.058-5.058 = 1.00
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