A 5.079 g sample of a solid, weak, monoprotic acid is used to make a 100.0 mL so
ID: 956950 • Letter: A
Question
A 5.079 g sample of a solid, weak, monoprotic acid is used to make a 100.0 mL solution. 26.00 mL of the resulting acid solution is then titrated with 0.09454 M NaOH. The pH after the addition of 18.00 mL of the base is 5.54, and the endpoint is reached after the addition of 47.60 mL of the base. Please see Titration to Determine Molecular Weight for assistance.
(a) How many moles of acid were present in the 26.00 mL sample? WebAssign will check your answer for the correct number of significant figures. mol
(b) What is the molar mass of the acid? WebAssign will check your answer for the correct number of significant figures. g/mol
(c) What is the pKa of the acid?
Explanation / Answer
end point : millimoles of acid = millimoles of base
millimoles of base = 0.09454 x 47.60 = 4.5
millimoles of acid = 4.5
molarity x volume = 4.5
molarity x 26.00 = 4.5
molarity = 0.173 M
molarity = ( mass / molar mass ) x 1000 / 100
0.173 = (5.079 / molar mass) x 1000 / 100
molar mass = 293 g/mol
millimoles = 4.5
moles of acid = 4.5 x 10^-3
millimoles of base = 0.09454 x 18 = 1.7
millimoles of acid = 26 x 0.173 = 4.50
acid + base ----------------> salt
4.50 1.7 0
2.8 0 1.7
pH = pKa + log [salt /acid]
5.54 = pKa + log (1.7/2.8)
pKa = 5.76
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