12) A student conducting a titration of a mono-protic weak acid by a mono-protic
ID: 103398 • Letter: 1
Question
12) A student conducting a titration of a mono-protic weak acid by a mono-protic strong base collected the data shown in the graph below. pH Titration Results 13.0 12.0 11.0 10.0 9.0 8.0 7.0 6.0 5.0 4.0 3.0 0.010 0.050 0.020 Vol. Added Base (L) 0.000 0.030 0.040 a) If the concentration of the strong base is 0.075 M, how many moles of acid were present in the sample? b) If the sample weighed 0.375 g, what is the formula weight of the unknown acid? c) Using graphical methods determine the pKa of the unknown acid. d) Calculate the pK, of the unknown acid using the initial pH value. The unknown acid was dissolved in enough solvent to make a solution with a volume of 10 mL. Show work.Explanation / Answer
a)
Answer
0.0021mol
Explanation
From graph , Equivalence point = 28ml
Concentration of Base = 0.075M
no of mole of Base = (0.075mol/1000ml)×28ml = 0.0021mol
0.0021mole of monoprotic base react with 0.0021mole of monoprotic acid, so
no of mole of monoprotic acid = 0.0021mol
b)
Answer
178.57g/mol
Explanation
Formula weight = Mass/no of mole
Formula weight of acid = 0.375g/0.0021mol
= 178.57g/mol
c)
Answer
6.0
Explanation
at half equivalence point
pH = pKa
half equivalence point = 0.014L
from graph, pH corresponding to 0.014L is 5.90
d)
Answer
6.12
Explanation
From graph , initial pH = 3.40
-log[H+] = 3.40
[H+] = 3.98×10^-4M
dissociation of monoprotic acid is
HA <------> H+ + A-
Ka = [H+] [A-]/[HA]
[H+] = [A-]
[HA] = 0.21 - 0.000398
Ka = (0.000398)^2/0.2096
= 7.56×10^-7
pKa = 6.12
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