Question 13. 76. Calculate the concentration of hydrogen ion in it. Answer: pH = – log_ten [H₃O + ] = – log₁₀ = – pH = – 3.76 = \(\overline\).dos4 [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 https://datingranking.net/escort-directory/plano/ 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. New pH out of 0.1 Meters solution out of cyanic acid (HCNO) try 2.34. Calculate new ionization ongoing of the acid as well as degree of ionization in the solution. HCNO \(\rightleftharpoons\) H + + CNO – pH = dos.34 setting – journal [H + ] = 2.34 or record [H + ] = – 2.34 = step three.86 otherwise [H + ] = Antilog 3.86 = 4.57 x ten -step three Yards [CNO – ] = [H + ] = 4.57 x ten -step 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.thirty-six x 10 -5 = 944 x 10 -eight pOH = – record (9.49 x 10 -seven ) = eight – 0.9750 = six.03 pH = fourteen – pOH = 14 – six.03 = seven.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, for dissolving 0.441 g, water required = I L For dissolving 1g, water required = \(\frac < 1>< 0.411>\)L = 2.43L

New solubility harmony throughout the saturated option would be AgCl (s) \(\rightleftharpoons\) Ag + (aq) + Cl – (aq) The solubility away from AgCl try step 1

1. Highlight the distinctions between ionic tool and solubility tool.
2. The fresh solubllity off AgCI in the water in the 298 K try step 1.06 x ten -5 mole per litre. Calculate try solubility product at that temperatures.

The fresh new solubility harmony in the saturated solution is AgCl (s) \(\rightleftharpoons\) Ag + (aq) + Cl – (aq) The fresh solubility from AgCl try step 1

1. It is relevant to all or any sort of options.
2. Their worth changes toward improvement in scam centration of your own ions.

The latest solubility harmony regarding the saturated solution is AgCl (s) \(\rightleftharpoons\) Ag + (aq) + Cl – (aq) Brand new solubility of AgCl are step one

1. It is appropriate on over loaded options.
2. It’s got a particular well worth for a keen electrolyte at a steady temperature.

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: