NCERT Class XI Chemistry Equilibrium Solutions

Question : 49

Total: 73

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

Solution:

(a) Molar mass of TlOH = 221.4 gmol–1
∴ nTlOH =

221.4gmol−1

= 9.03 × 10−3‌mol
[OH−] = [TlOH] =

9.03×10−3‌mol

= 4.51 × 10−3M
pOH = – log(4.51 × 10–3) = 2.35 and pH = 14 – 2.35 = 11.65
(b) nCa(OH)2 =

0.3g

74gmol−1

= 4.05 × 10−3‌mol
[OH−] = 2[Ca(OH)2] = 2 ×

4.05×10−3‌mol

0.5L

= 1.62 × 10−2M
pOH = - log (1.62 × 10−2) = 1.79 and pH = 14 - 1.79 = 12.21
(c) nNaOH =

0.3g

40gmol−1

= 7.5 × 10−3‌mol
[OH−] = [NaOH] =

7.5×10−3‌mol

0.2L

= 0.0375 M
pOH = – log(0.0375) = 1.43 and pH = 14 – 1.43 = 12.57
(d) M1V1 = M2V2
13.6 mol‌L–1 × 1 mL = M × 1000 mL
MHCl = 0.0136 mol‌L–1
[H+] = [HCl] = 0.0136 M
∴ pH = – log(0.0136) = 1.87

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