A buffer solution was prepared by dissolving 5.00 grams of sodium propanate (NaC
ID: 793284 • Letter: A
Question
A buffer solution was prepared by dissolving 5.00 grams of sodium propanate (NaCH3CH2CO2) in a solution containing 0.100 moles of propanoic acid (CH3CH2CO2H) and diluting the mixture to 500.0 mL. To this solution was added 5.00 mL of 1.00 M HCl. Calculate the pH of the resulting solution. pKa for propanoic acid is 4.874
A buffer solution was prepared by dissolving 5.00 grams of sodium propanate (NaCH3CH2CO2) in a solution containing 0.100 moles of propanoic acid (CH3CH2CO2H) and diluting the mixture to 500.0 mL. To this solution was added 5.00 mL of 1.00 M HCl. Calculate the pH of the resulting solution. pKa for propanoic acid is 4.874Explanation / Answer
sod propanoate moles = 5/96.07 = 0.052
moles of propanoic acid = 0.1
HCl moles added = 1 x5/1000 = 0.005
Hence total propanoic acid moles = 0.1+0.005 = 0.1005
sodium propanoate moles = 0.052-0.005 = 0.047
pH = pka + log [sodium propanoate]/[propoanoic acdi]
pH = 4.874 + log ( 0.047/0.55/0.1005/0.55)
pH = 4.544
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