Question:
Calculate the pH of the following solutions:

(a) 2 g of TlOH dissolved in water to give 2 litre of solution.

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

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

(d) 1mL of 13.6 M HCl is diluted with water to give 1 litre of solution.

Answer:

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

[TIOH(aq)] = 2/2 g/L

= 2/2 x 1/221 M

= 1/221 M

TIOH_(aq) → TI⁺_(aq) + OH^-_(aq)

OH^-_(aq)= TIOH_(aq)= 1/221M

Kw = [H⁺] [OH^-]

10^-14= [H⁺] [1/221]

[H⁺] = 221x10^-14

⇒ pH = -log [H⁺] = -log ( 221x10^-14)

= 11.65

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

Ca(OH)₂ → Ca²⁺ + 2OH^-

[Ca(OH)₂] = 0.3x1000/500 = 0.6M

OH^-_(aq) = 2 x [Ca(OH)_2(aq)] = 2 x 0.6 = 1.2M

[H+] = Kw / OH^-_(aq)

= 10^-14/1.2 M

= 0.833 x 10^-14

pH = -log(0.833 x 10^-14)

= -log(8.33 x 10^-13)

= (-0.902 + 13)

= 12.098

NaOH → Na⁺_(aq) + OH^-_(aq)

[NaOH] = 0.3 x 1000/200 = 1.5M

[OH^-_(aq)] = 1.5M

Then [H+] = 10^-14/ 1.5

= 6.66 x 10^-13

pH = -log ( 6.66 x 10^-13)

= 12.18

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

13.6 x 1 mL = M₂x 1000 mL

(Before dilution) (after dilution)

13.6 x 10^-3 = M₂ x 1L

M₂ = 1.36 x 10^-2

[H⁺] = 1.36 × 10^-2

pH = - log (1.36 × 10^-2)

= (- 0.1335 + 2)

= 1.866 = 1.87

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