Question
In: Chemistry
A 1.00 L buffer solution is 0.150 M CH3COOH (pKa = 4.74 ) and 0.200 M...
A 1.00 L buffer solution is 0.150 M CH3COOH (pKa = 4.74 ) and 0.200 M NaCH3COO. What is the pH of this buffer solution after 0.0500 mol KOH are added?
Solutions
Expert Solution
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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