7.49. Calculate the pH of the following solutions:
(a) 2g ofTlOH dissolved in water to give 2 litre of the solution
(b) 0.3 g of Ca(OH)2 dissolved in water to give 500 mL of the solution
(c) 0.3 g of NaOH dissolved in water to give 200 mL of the solution
(d) l mL of 13.6 M HCl is diluted with water to give 1 litre of the solution.
7.49. Calculate the pH of the following solutions:
(a) 2g ofTlOH dissolved in water to give 2 litre of the solution
(b) 0.3 g of Ca(OH)2 dissolved in water to give 500 mL of the solution
(c) 0.3 g of NaOH dissolved in water to give 200 mL of the solution
(d) l mL of 13.6 M HCl is diluted with water to give 1 litre of the solution.
(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
Similar Questions for you
0.01 M NaOH,
M = 1 * 10-2

pOH = 2
pH = 2
Kp = Kc (RT)Dng
36 * 10–2 = Kc (0.0821 * 300)–1
Kc = 0.36 * 0.0821 * 300 = 8.86 » 9
A(g) ->B(g) + (g)
Initial moles n 0 &nbs
On increasing pressure, equilibrium moves in that direction where number of gaseous moles decreases.
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Chemistry Ncert Solutions Class 11th 2023
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