Calculate the pH of the following soluti | Class 11 Chemistry Chapter Equilibrium, Equilibrium NCERT Solutions

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

 

(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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Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 49: Calculate the pH of the following solutions: (a) 2 g of TlOH dissolved in water to give 2 litre o....