Calculate pH for each of the following cases in the titration of 25 mL of 0.140
ID: 699727 • Letter: C
Question
Calculate pH for each of the following cases in the titration of 25 mL of 0.140 M pyridine (aq) with 0.140 M HBr (aq)
a) before addition of any HBr
Pyridine is a weak base and reacts with water to generate –OH ion as shown below:
C5H5N + H2O ç=====è C5H5NH+ + –OH
Which may be rewritten as follows:
Py + H2O ç=====è PyH+ + –OH
Accordingly the equilibrium constant for the above reaction can be written as
Kb = [Py][ –OH]/[Py]
Note: Concentration of H2O is not included in Kb calculation as it is in excess and any small change in its concentration doesn’t influence the equilibrium.
Now for calculating Kb we have to generate an ICE (Initial, Change in concentration, Equilibrium concentration) table to understand the changes during reaction as shown below:
Reaction
Py , [M]
PyH+, [M]
OH [M]
Initial concentration
0.14
0
0
Change in concentration
-x
+x
+x
Equilibrium concentration
0.14-x
x
x
From literature we know Kb = 1.7 x 10-9
Therefore if we plug the above values into Kb formula we get
1.7 x 10-9 = [x][x]/[0.14-x]
To simplify the calculation ‘x’ value in denominator is omitted due to its small value in comparison to 0.14M. Hence we get as follows
x2 = [1.7 x 10-9 ][0.14]
x2 = 2.38 x 10-10
x = Square root of 2.38 x 10-10
x = 1.543 x 10-5
Therefore concentration of –OH will be equal to x value at the equilibrium. Therefore
[–OH] = 1.543 x 10-5 M
Since
pOH = - log [–OH]
Therefore
pOH = - log [1.543 x 10-5]
pOH = 4.81
Since
pH = 14 – pOH
Therefore
pH = 14 – 4.81 = 9.19
Therefore pH of solution before addition of any HBr is 9.19
b) after addition of 12.5 mL of 0.14 M HBr
As pyridine reacts with aqueous acid added as shown below
Py + H3O+ ç=======è PyH+ + H2O
Therefore the solution behaves as a buffer and pOH can be calculated from Henderson Hasselbach equation as shown below
pOH = pKb + log [PyH+]/[Py]
Initial moles of pyridine = 0.14 x 0.025 = 0.0035 Moles
Since pyridine reacts with HBr and is neutralized by the amount of acid added
After addition of 12.5 mL of 0.14 M HBr moles, the same amount of pyridine will react with HBr and the remaining amount of pyridine will be equal to
Amount of HBr added = 0.14 x 0.0125 = 0.00175 moles
remaining amount of pyridine = 0.0035 – 0.00175 = 0.00175 moles
Therefore concentration of remaining pyridine = number moles of pyridine/total volume of solution
Concentration of remaining pyridine = 0.00175 / [0.025+0.0125]
= 0.00175 / [0.025+0.0125]
= 0.00175/0.0375
= 0.046 M
Concentration of remaining pyridine = 0.046 M
Concentration of PyH+ generated will be equal to 0.046 M
Therefore
Since pKb = -log Kb = 1.7 x 10-9 = 8.77
pOH = 8.77 + log [0.046/0.046]
pOH = 8.77 + 0
pOH = 8.77
pH = 14 – 8.77
pH = 5.23
After addition of 12.5 mL of 0.14 M HBr the pH of solution is 5.23
c) after addition of 14.0 mL of HBr
Amount of HBr added = 0.14 x 0.014 = 0.00196 moles
remaining amount of pyridine = 0.0035 – 0.00196 = 0.00154 moles
Concentration of PyH+ generated will be equal to acid added = 0.00196 moles
pOH = 8.77 + log [0.00196/0.00154]
pOH = 8.77 + log 1.27
pOH = 8.77 + 0.10
pOH = 8.87
pH = 14 – 8.87 = 5.13
After addition of 14 mL of 0.14 M HBr the pH of solution is 5.13
d) after addition of 25.0 mL of HBr
Since the concentration of initial pyridine and added HBr is same and also the volumes are also equal the solution will be neutral and its pH will be equal to 7.0.
e) after addition of 35.0 mL of HBr
Reaction
Py , [M]
PyH+, [M]
OH [M]
Initial concentration
0.14
0
0
Change in concentration
-x
+x
+x
Equilibrium concentration
0.14-x
x
x
Explanation / Answer
Question 25 of 27 Map General Chemistry 4th Edition McQuarrie University Science Books presented by Sapling Learning Rock Gallo Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.140 M pyridine, C5H5N(aq) with 0.140 M HBr(aq): Number (a) before addition of any HBr Number (b) after addition of 12.5 mL of HBrD Number (c) after addition of 14.0 m of HBr Number (d) after addition of 25.0 mL of HBr Number (e) after addition of 35.0 mL of HBr
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