Question

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

Answer 1

1.Ans :-

Given K_a of HNO₂ = 4.5 x 10^-4

K_b of NO₂^- = 1.0 x 10^-14 / K_a

= 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. K_b(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).

K_b = [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 = pK_a + log [Conjugate base] / [Acid]

pH = - log K_a​​​​​​​  + 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. K_a(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).

K_a = [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