Equilibrium

Let c be the initial concentration of C₆H₅NH₃⁺ and x be the degree of ionisation.

C₆H₅NH₂ + H₂O?  C₆H₅NH₃⁺ + OH^-

c (1-x)                      cx              cx

K_b = [C₆H₅NH₃⁺] [ OH^-] / [C₆H₅NH₂]

= [cx] [cx] / [c (1 – x)]

Since x is very small and negligible 1 – x≈ 1

 ∴K_b= [cx] [cx] / [c] = cx²

=> x = $\sqrt[]{\frac{Kb}{c}}$

=

= 6.56 x 10^-4

∴ [OH^-] = cx = 0.001 x 6.56 x 10^-4 = 6.56 x 10^-7 M

  [H⁺]= Kw / [OH^-] = 10^-14 / 6.56 x 10^-7 = 1.52 x 10^-8

pH= –log [H⁺] = –log1.52 x 10^-8 = 7.818

∴ Ionization constant of the conjugate acid of aniline,

K_a = K_w / K_b

= 10^-14 / (4.3 x 10^-10)

= 2.32 x 10^-5

(a) For 2g of TlOH dissolved in water to give 2 L of solution:

[TlOH] = [OH⁻] = (2*1)? / (2*221) = (1 / 221)? M

pOH = −log [OH]⁻ = −log (1/221)?

= 2.35

pH = 14 – pOH = 14 − 2.35 = 11.65

(b) For 0.3 g of Ca (OH)2?  dissolved in water to give 500 mL of solution:

[OH⁻] = 2 [Ca (OH)₂? ] = 2 (0.3*1000/500? ) = 1.2M
pOH = −log [OH⁻] = −log1.2 = 1.79

pH= 14−pOH=14−1.79

=12.21

(c) For 0.3 g of NaOH dissolved in water to give 200 mL of solution:

[OH⁻]= [NaOH] = 0.3*1000/200? = 1.5M
pOH= −log [OH⁻] = −log1.5 = 1.43

pH= 14 – pOH = 14 − 1.43

= 12.57

(d) For 1mL of 13.6 M HCl diluted with water to give 1 L of solution:

The molarity of HCl solution after dilution is  (13.6*1)/ 1000= 0.0136M.
It is equal to hydrogen ion concentration.
pH = −log [H⁺]= −log0.0136

= 1.87

The dissociation equilibrium is 

CH_3? COOH? CH3? COO⁻ + H⁺.
Let α be the degree of dissociation.
The equilibrium concentrations of CH_3? COOH, CH₃COO⁻ and H⁺ are c (1−α), c (α) and c (α) respectively.
The equilibrium constant expression is K_c? = [CH_3? COO⁻] [H⁺]? / [CH₃? COOH].

K_c? = (cα) (cα) / c (1−α)? ≈cα²

α= (K_a? / c)^1/2? = (1.74*10⁻⁵ / 0.05)^1/2?

=1.865*10⁻²
[CH3? CO⁻]= [H⁺]= cα= 0.05*1.865*10⁻²= 9.33*10⁻⁴M
pH= −log [H⁺]= −log (9.33*10⁻⁴)= 3.03

The concentration of acetate ion and its pH are 9.33*10⁻⁴ and 3.03 respectively.

(i) To calculate  [HS⁻] in absence of HCl:
Let,   [HS⁻] = x M.

H_2? S? H⁺ + HS⁻
The initial concentrations of H_2? S, H⁺ and HS⁻ are 0.1 M, 0 M and 0 M respectively.
Their final concentrations are 0.1-x M, x M and x M respectively.

K_a? = [H₂? S] [H⁺] [HS⁻]?
  9.1*10⁻⁸ = x * x / (0.1−x)?
In the denominator, 0.1-x can be approximated to 0.1 as x is very small.
   9.1*10⁻⁸=x*x / (0.1)?

x=9.54*10⁻⁵M= [HS⁻]

(ii) To calculate  [HS⁻] in presence of HCl:
Let  [HS⁻]= y M.

 H₂? S? H⁺+HS⁻
The initial concentrations of H_2? S, H⁺ and HS⁻ are 0.1 M, 0 M and 0 M respectively.
Their final concentrations are 0.1-y M, y M and y M respectively.
Also, HCl? H⁺+Cl⁻
For HCl  [H⁺]= [Cl⁻]=0.1M
K_a? = [HS⁻] [H⁺] / [H_2? S]?
K_a? = y (0.1+y)? / (0.1−y)
In the denominator, 0.1-y can be approximated to 0.1 and in the numerator, 0.1+y can be approximated to 0.1 as y is very small.
9.1*10⁻⁸= (y*0.1) / 0.1?

y= 9.1*10⁻⁸M= [HS⁻]

(iii) To calculate [S²⁻] in absence of 0.1 M HCl:

HS⁻? H⁺+S²⁻
[HS⁻]=9.54*10⁻⁵M
Let  [S²⁻]= x M.
[H⁺]= 9.54*10⁻⁵ M

K_a? = [H⁺] [S²⁻]? / [HS⁻]
= >1.2*10⁻¹³= (9.54*10⁻⁵*X? ) / (9.54*10⁻⁵)
= > x=1.2*10⁻¹³M= [S²⁻]

(iv) To calculate  [S²⁻] in presence of 0.1 M HCl:
Let  [S²⁻]= x′ M.
[HS⁻]= 9.1*10⁻⁸ M
[H⁺]= 0.1 M (from HCl)

K_a? = [HS⁻] [H⁺] [S²⁻]?
1.2*10⁻¹³= (0.1*X′? ) / (9.1*10⁻⁸)

x′= 1.092*10⁻¹⁹= [S²⁻]