# pH OF Weak Acids and Bases:

Weak acids and bases are not completely ionised; an equilibrium is found to have been established between ions and unionised molecule. Let us consider a weak acid of basicity 'n'.

AHn  ↔  An-  +   nH+

i  = 0    -C        0          0

teq      C(1-α)   Cα       nCα

[H+] = nCα;  .·. pH = -log10 [nCα]                ...... (i)

For monobasic and,  n=1

pH = -log10 [Cα]                                       ...... (ii)

Dissociation constant of acid Ka may be calculated as

Ka = [An-][H+]n/[AHn] = [Cα][nCα]n/[C(1-α)]

=  α [nCα]n/(1-α)           For weak acids, α« 1

.·. (1-α) = 1

= α[nCα ]n/(1-α)

nCKa = nCα [nCα ]n

= [nCα ](n+1)

= [nCα ] = [nCKa]1/(n+1)

= [H+] = [nCKa]1/(n+1)

.·. pH = -1/(n+1) log10(nCKa)              ...... (iii)

For monobasic acid, n = 1

pH = -log√CKα                                ....... (iv)

Since Ka = α[nCα]n

ka/α = (nCα)n

[nCα ] = [Kα/α]1/n = [H+]

pH = -1/n log10(Kα/α)                          ..... (v)

For n = 1       pH = -log10(Kα/α)          ..... (vi)

Example 28:  The hydrogen ion concentration of a solution is 0.001 M. What will be the hydroxyl ion concentration of solution?

Solution:       We know that [H+][OH-] 1.0 × 10-14

Given that,     [H+] = 0.001 M = 10-3 M

So                [OH-] = 1.0 * 10-14/[H+] = (1* 10-14)/10-3 = 10-11M

Example 29:  What is the pH of the following solutions?

(a) 10-3 M HCl

(b) 0.0001 M NaOH

(c)  0.0001 MH2S04

Solution:   HCI is a strong electrolyte and is completely ionised.

HCl ↔ H+ + CI-

So   [H+] = 10-3 M

pH = -log [H+] = -log (10-3) = 3

(b)   NaOH is a strong electrolyte and is completely ionised.

NaOH ↔ Na+ + OH-
So    [OH+] = 0.0001 M = 10-4 M

pOH = -log(10-4) = 4

As   pH + pOH = 14

So   pH + 4 = 14   or     pH=10

Alternative method:   [OH-] = 10-4 M

We know that    [H+][OH-] = 1.0 x 10-14

So  [H+] = (1.0 * 10-14)/10-14

pH = -log [H+] = -log(10-10) = 10

(c) H2SO4 is a strong electrolyte and is ionized completely.

H2SO4 ↔ 2H+ + SO2-4

One molecule of H2SO4 furnishes ions.

So [H+] = 2 × 10-4 M

pH = -log [H+]

= -log (2×10-4)

= 3.70

Example 30:  Calculate the pH of the following solutions assuming complete dissociation:

(a) 0.365 g L-1 HCl solution

(b) 0.001 M Ba(OH)2 solution.

Solution: (a) Mole, mass of HCl = 36.5

Concentration of HCl = 0.365/36.5 = 1.0 × 10-2 mol L-1

HCl is a strong electrolyte and is completely ionised.

So     [H+] = 1 × 10-2 mol L-1

pH c -log [H+] = -log (1 × 10-2) = 2

(b) Ba(OH)2 is a strong electrolyte and is completely ionised

Ba(OH)2 ↔ Ba2+ + 2OH-

One molecule on dissociation furnishes 2 OH- ions.

So     [OH-] = 2 × 10-3 M

pOH = -log [OH-]

= -log (2 × 10-3) = 2.7

We know that pH + pOH = 14

So     pH = (14-2.7) = 11.3

Example 31: Find the pH of a 0.002 N acetic acid solution, if it is 2.3% ionised at a given dilution.

Solution: Degree of dissociation, α = 2.3/100 = 0.023

Concentration of acetic acid, C = 0.002 M

The equilibrium is

CH3COOH ↔ CH3COO- + H+

C(1-α)         Cα           Cα

So  [H+] = Cα  = 0.002 × 0.023

= 4.6 × 10-5 M

pH = -log [H+]

= -log (4.6 × 10-5) = 4.3372

Example 32:   Calculate the pH value of a solution obtained by mixing 50 mh of 0.2 N HCl with 50 mL of 0.1 N NaOH.

Solution:    Number of milli-equivalents of the acid

= 50 × 0.2 = 10

Number of milli-equivalents of the base

= 50 × 0.1 =5

Number of milli-equivalents of the acid left after the addition of base

= (10-5) = 5 Total volume of the solution

= 50 + 50 = 100 mL

Thus, 5 milli-equivalents of the acid are present in 100 mL of solution.

or 50 milli-equivalents of the acid are present in one litre of solution.

or 0.05 equivalents of the acid are present in one litre of solution.

The acid is monobasic and completely ionised in solution.

0.05 N HCl = 0.05 M HCl

So     [H+] = 0.05 M

pH = -log [H+] = -log 5 × 10-2 = -[log 5.0 + log 10-2]

= -[0.70-2] = 1.3

Example 33 :  What will be the pH of a solution obtained by mixing 800 mL of 0.05 A' sodium hydroxide and 200 mL of OA N HCl, assuming complete ionisation of the acid and the base ?

Solution:   Number of milli-equivalents of NaOH

= 800 × 0.05 = 40

Number of milli-equivalents of HCl

= 200 × 0.1 =20

Number of milli-equivalents of NaOH left after the addition of HCl

= (40 -20) = 20

Total volume = (200 + 800) mL = 1000 mL = 1 litre

20 milli-equivalents or 0.02 equivalents of NaOH are present in one litre, i.e.,

0.02 N NaOH = 0.02 M NaOH (Mono-acidic) and the base is completely ionised.

So     [OH-] = 0.02 M

or     [OH-]= 2 × 10-2 M

pOH = -log(2 × 10-2) = 1.7

We know that,   pH + pOH = 14

So   pH = (14-1.7)= 12.3

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