nitrous acid pKa 4.5E10-4; HF 7.1E-4; Formic Acid 1.7E-4; Acetic Acid 1.8E-5; HCH 4.9E-10
1 Consider the titration of 40.0 mL of 0.0500 M NO2- with 0.100 M HBr
a calculate the pH of the solution before the addition of any HBr
b calculate the pH of the solution after the addition of 10.0 mL of HBr
c calculate the pH of the solution after the addition of 20.0 mL of HBr
1.Ans :-
Given Ka of HNO2 = 4.5 x 10-4
Kb of NO2- = 1.0 x 10-14 / Ka
= 1.0 x 10-14 / 4.5 x 10-4
= 2.2 x 10-11
(a). pH of the solution before the addition of any HBr :-
ICE table of NO2- is :
............................NO2- (aq)............+...............H2O (l) <-----------------> HNO2 (aq)............+.................OH- (aq)
Initial................0.0500 M.........................................................................0 M..........................................0 M
Change..............-y...................................................................................+y..............................................+y
Equilibrium..........(0.0500-y) M...............................................................y M...........................................y M
Where,
y = Amount dissociated per mole
Expression of equilibrium constant i.e. Kb(which is equal to the product of the molar concentration of products divided by product of the molar concentration of reactants raise to power their stoichiometric coefficient when reaction is at equilibrium stage).
Kb = [HNO2].[OH-] / [NO2-]
2.2 x 10-11 = y2 / (0.0500-y)
As y <<<0.0500, then neglect y as compare to 0.0500
So,
y2 = 2.2 x 10-11 x 0.0500
y = (1.1 x 10-12)1/2
y = 1.05 x 10-6
So, [OH-] = y = 1.05 x 10-6 M
pOH = -log [OH-]
= - log 1.05 x 10-6 M
= 5.98
pH = 14 - pOH
pH = 14 - 5.98
pH = 8.02
Therefore, pH = 8.02 |
(b). pH of the solution after the addition of 10.0 mL of HBr :-
No. of moles of NO2- = Molarity x volume in L
= 0.0500 M x 0.040 L
= 0.002 mol
No. of moles of HBr = Molarity x volume in L
= 0.100 M x 0.010 L
= 0.001 mol
Now, ICF table is :
................................NO2- ..............+................HBr ----------------> HNO2 ................+....................Br-
Initial......................0.002 mol..........................0.001 mol ..................0 mol
Change...............-0.001 mol .........................-0.001 mol .............+0.001 mol
Final......................0.001 mol .............................0 mol......................0.001 mol
From Henderson-Hasselbalch equation :
pH = pKa + log [Conjugate base] / [Acid]
pH = - log Ka + log [NO2-] / [HNO2]
pH = - log 4.5 x 10-4 + log 0.001 / 0.001
pH = 3.35 + log 1
pH = 3.35 + 0
pH = 3.35
Therefore, pH = 3.35 |
(c). pH of the solution after the addition of 20.0 mL of HBr :-
No. of moles of NO2- = Molarity x volume in L
= 0.0500 M x 0.040 L
= 0.002 mol
No. of moles of HBr = Molarity x volume in L
= 0.100 M x 0.020 L
= 0.002 mol
Now, ICF table is :
................................NO2- ..............+................HBr ----------------> HNO2 ................+....................Br-
Initial......................0.002 mol..........................0.002 mol ..................0 mol
Change...............-0.002 mol .........................-0.002mol .............+0.002 mol
Final......................0 mol ....................................0 mol......................0.002 mol
Total volume = 40 mL + 20 mL = 60 mL = 0.060 L
[HNO2] = Moles / Volume in L
= 0.002 mol / 0.060 L
= 0.033 M
So,
ICE table of HNO2 is :
............................HNO2 (aq) <--------------> H+ (aq)............+.................NO2- (aq)
Initial................0.033 M............................0 M..........................................0 M
Change..............-y...................................+y..............................................+y
Equilibrium..........(0.033-y) M...............y M...........................................y M
Where,
y = Amount dissociated per mole
Expression of equilibrium constant i.e. Ka(which is equal to the product of the molar concentration of products divided by product of the molar concentration of reactants raise to power their stoichiometric coefficient when reaction is at equilibrium stage).
Ka = [NO2-].[H+] / [HNO2]
4.5 x 10-4 = y2 / (0.033-y)
y2 + 4.5 x 10-4 y - 1.485 x 10-5 = 0
On solving
y = 0.00364 M = [H+]
pH = - log [H+]
pH = - log 0.00364 M
pH = 2.44
Therefore, pH = 2.44 |
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