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A 1.00 L buffer solution is 0.150 M CH3COOH (pKa = 4.74 ) and 0.200 M...

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Chemistry

A 1.00 L buffer solution is 0.150 M CH3COOH (pKa = 4.74 ) and0.200 M NaCH3COO. What is the pH of this buffer solution after0.0500 mol KOH are added?

Answer & Explanation Solved by verified expert
3.8 Ratings (374 Votes)

no of moles of CH3COOH = molarity*volume in L

                                            = 0.15*1 = 0.15 moles

no of moles of CH3COONa = molarity*volume in L

                                            = 0.2*1 = 0.2 moles

PH   = PKa + log[CH3COONa]/[CH3COOH]

        = 4.74 + log0.2/0.15

      = 4.74 + 0.1249   = 4.8649

By the addition of 0.05 moles of KOH

no of moles of CH3COOH = 0.15-0.05 = 0.1 moles

no of moles of CH3COONa   = 0.2+0.05 = 0.25 moles

PH   = PKa + log[CH3COONa]/[CH3COOH]

        = 4.74 + log0.25/0.1

       = 4.74 + 0.3979   = 5.1379
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