Class 11 Chemistry - Chapter Equilibrium NCERT Solutions | Calculate the pH of the following soluti
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.
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
(c) For 0.3 g of NaOH dissolved in water to give 200 mL of solution:
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 = M2 x 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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