Question 13. 76. Calculate the concentration of hydrogen ion in it. Answer: pH = – log_ten [H₃O + ] = – log₁₀ = – pH = – 3.76 = \(\overline\).24 [H₃O + ] = antilog \(\overline<4>\).24 = l.738 x 10 -4 [H₃O + ] = 1.74.x 10 -4 M

Question 14. The ionisation constant of HF, HCOOH, HCN at 298 K are 6.8 x 10 -4 , 1.8 x 10 -4 and 4.8 x 10 -9 respectively. Calculate the ionisation constant of the corresponding conjugate base. Answer: 1. HF, conjugate base is F K_b = K_w/K_a = \(\frac<1>><6.8>>\) = l.47 x 10 -11 = l.5 x 10 -11

Matter fifteen. The fresh pH of 0.step 1 Yards provider away from cyanic acidic (HCNO) is dos.34. Estimate the latest ionization constant of your acid and its particular amount of ionization in the solution. HCNO \(\rightleftharpoons\) H + + CNO – pH = dos.34 function – log [H + ] = dos.34 otherwise record [H + ] = – 2.34 = step 3.86 or [H + ] = Antilog step three.86 = 4.57 x 10 -3 Yards [CNO – ] = [H + ] = cuatro.57 x ten -3 Meters

Question 16. The Ionization constant of nitrous acid is 4.5 x 10 -4 . Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis. Answer: Sodium mtrite is a salt of weak acid, strong base. Hence, K_h = 2.22 x 10_-11 K_w/K_b = 10 -14 /(4.5x 10 -4 )

[OH – ] = ch = 0.04 x dos.36 x 10 -5 = 944 x 10 -seven pOH = – log (9.49 x ten -7 ) = 7 – 0.9750 = six.03 pH = 14 – pOH = fourteen – six.03 = eight.97

Question 17. What is the minimum volume of water required to dissolve 1 g of calcium sulphate at 298K. For calcium sulphate, K_sp = 9.1 x 10 -6 . Answer: CaSO₄(s) Ca 2 (aq) + SO 2- ₄(aq) If ‘s’ is the solubility of CaSO₄ in moles L – , then K_sp = [Ca 2+ ] x [SO₄ 2- ] = s 2 or

= 3.02 x 10 -3 x 136gL -1 = 0.411 gL -1 (Molar mass of CaSO₄ = 136 g mol -1 ) Thus https://datingranking.net/escort-directory/new-orleans/, for dissolving 0.441 g, water required = I L For dissolving 1g, water required = \(\frac < 1>< 0.411>\)L = 2.43L

New solubility equilibrium in the soaked solution is AgCl (s) \(\rightleftharpoons\) Ag + (aq) + Cl – (aq) The brand new solubility from AgCl is actually step one

1. Suggest the differences between ionic unit and solubility product.
2. The solubllity regarding AgCI in the water at the 298 K try 1.06 x ten -5 mole for every litre. Assess was solubility tool at this heat.

The solubility harmony regarding the over loaded solution is AgCl (s) \(\rightleftharpoons\) Ag + (aq) + Cl – (aq) The new solubility out of AgCl try step 1

1. It’s relevant to any or all style of alternatives.
2. Their value alter into improvement in con centration of your own ions.

The latest solubility harmony about over loaded solution is AgCl (s) \(\rightleftharpoons\) Ag + (aq) + Cl – (aq) This new solubility away from AgCl is actually step 1

1. It’s appropriate with the saturated alternatives.
2. It’s got one value to have an enthusiastic electrolyte within a stable temperatures.

2. 06 x 10 -5 mole per litre. [Ag + (aq)] = 1.06 x 10 -5 mol L -1 [Cl – (aq)] = 1.06 x 10 -5 mol L -1 K_sp = [Ag + (aq)] [Cl – (aq)] = (1.06 x 10 -5 mol L -1 ) x (1.06 x 10 -5 mol L -1 ) = 1.12 x 10 -2 moI 2 L -2

Question 19. The value of K of two sparingly soluble salts Ni(OH)₂ and AgCN are 2.0 x 10 -15 and 6 x 10 -17 respectively. Which salt is more soluble? Explain. Answer: