please show me answer and explanation with good hand wrting! Question B1 Morphol
ID: 560766 • Letter: P
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please show me answer and explanation with good hand wrting!
Question B1 Morpholine (C4H9NO, 87.12 g.mor1) is a weak base with K.-3.09 × 10-6. A sample weighing 2.51 g was dissolved in enough water to make 250.00 mL of solution, and a 50.00 mL portion of this solution was then titrated with 0.150 M HCl. (a) What volume of HCl is required to reach the equivalence point?[1 mark] (b) Determine the pH of the solution: (i) Before any HCl is added 1 markj (i) After exactly 25.00 mL of HCl is added [1 mark] (ii) At the equivalence point [1 mark]Explanation / Answer
From the given data
molarity of morpholine = (mass/molar mass) /V in L
= (2.51g/87.12g/mol) /0.25L
= 0.1152 M
a) The volume of HCl at equivalence
AT equivalence
mmoles of base = mmoles of acid
50mL x 0.1152M = 0.15M x V
Thus volume of HCl = 38.41 at equivalence
b) pH before any HCl is added
The pH of weak base = 14-pOH
pOH = 1/2[pKb -logC]
= 1/2 [ 5.51 -log 0.1152]
= 3.224
thus pH = 14-3.224 = 10.776
b) after adding 25 mL of HCl
B + HCl -------------> BH+ + Cl-
0.1152x50=5.76 0.15x25=3.75 0 0 initial mmoles
2.01 0 3.75 - after reaction
Thus the solution is a buffer of weak base and its conjugate acid
pH is caluted by using Hendersen equation as
pH = 14 -[ pkb +log [conjugate acid]/[base]}
= 14 -[ 5.51 + log 3.75/2.01]
= 8.219
d)At equivalence point
B + HCl -------------> BH+ + Cl-
0.1152x50=5.76 0.15x38.41=5.76 0 0 initial mmoles
0 0 5.76 5.76 at equivalence
Thus the solution has only salt of weak base and strong acid abd its aqueous solution is acidic due to cationic hydrolysis.
The concentration of salt = mmoles/ total volume
= 5.76/ 88.41
Its pH is calculated as
pH = 1/2[pKw -pKb -log C]
= 1/2[ 14 -5.51 - log 5.76/88.41]
= 4.838
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