1-A 0.255 gram sample of a weak base is dissolved in 100.0 ml of distilled water
ID: 1001310 • Letter: 1
Question
1-A 0.255 gram sample of a weak base is dissolved in 100.0 ml of distilled water. A 20.0 ml sample of this solution is titrated with 0.100 M HCl. A colorimetric indicator, pKa = 5.1 and a pH meter were used to generate the graph. Equivalence was found to occur at 30.0 ml of HCl added.
a- Determine the molecular mass of the weak base. Show your calculations
b- Determine the pKb of the weak base. Justify your answer
c- If a pH indicator with a pKa = 9.1 was used by accident, how would the calculated molar mass be affected? Justify your answer
d- When the 20.0 ml sample is transferred to the titration flask, there was 7 ml of water in the flask. How would this affect the calculated molar mass? Justify your answer
e- Describe the contents of the flask, in terms of weak base and conjugate acid, at 0ml, 10ml, 15 ml and 30ml of HCl added.
f- Why is the pH at equivalence < 7? Explain
answer as soon as possible please!
Explanation / Answer
1. MaVa=MbVb
0.100 M * 20 ml = Mb * 100 ml
Mb = 0.100 M * 20 ml /100 ml = 0.02M
Molarity = number of moles / Volume in Litres = mass of base / Molar mass of base * Volume in Litres
0.02M = 0.255 g / Molecular mass of base * 100 ml * 0.001 L
Molecular mass of base = 0.255 g / 0.02 M * 100 ml * 0.001 L
Molecular mass of weak base = 127.5 g mol-1
2. pKb = 14- pKa
pKb = 14 - 5.7
pKb = 8.3
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