100cm3 buffer solution of 0.1M CH3COOH and 0.05M CH3COONa are added 3cm3 HCl 0.01M. Ph =?

Question

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Answer 1

Using moles and the assumption that the HCl will be fully neutralized by the A^- (acetate ion):

K_a = 1.8 x 10^-5 pK_a = 4.74

moles of HA = .1 mole/liter (.1 liter) = .01 moles

moles of A^- = .05 mole/liter (.1 liter) = .005 moles

moles of HCl = .003 liter (.01 mole/liter) = 3 x 10^-5 moles

New moles of HA = .01 + 3 x 10^-5

New moles of A^- = .005 - 3 x 10^-5

Total volume = .103 liters

Henderson-Hasselbalch for the system:

pH = 4.74 + log (.00497/.103)/(.01003/.103) = 4.43

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