Question 13. 76. Calculate the concentration of hydrogen ion in it. Answer: pH = – log_{ten} [H_{3}O + ] = – log_{10} = – pH = – 3.76 = \(\overline_{3}O + ] = antilog \(\overline<4>\).24 = l.738 x 10 -4 [H_{3}O + ] = 1.74.x 10 -4 M

Question 14. The ionisation constant of HF, HCOOH McAllen escort reviews, 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

Concern fifteen. New pH away from 0.1 M services regarding cyanic acidic (HCNO) are dos.34. Assess the new ionization lingering of your acidic as well as amount of ionization regarding solution. HCNO \(\rightleftharpoons\) H + + CNO – pH = 2.34 function – record [H + ] = dos.34 otherwise journal [H + ] = – 2.34 = step three.86 or [H + ] = Antilog step three.86 = cuatro.57 x ten -3 Yards [CNO – ] = [H + ] = 4.57 x 10 -3 Yards

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 2.thirty-six x 10 -5 = 944 x ten -eight pOH = – log (9.44 x ten -seven ) = 7 – 0.9750 = six.03 pH = 14 – pOH = fourteen – 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: CaSO_{4}(s) Ca 2 (aq) + SO 2- _{4}(aq) If ‘s’ is the solubility of CaSO_{4} in moles L – , then K_{sp} = [Ca 2+ ] x [SO_{4} 2- ] = s 2 or

= 3.02 x 10 -3 x 136gL -1 = 0.411 gL -1 (Molar mass of CaSO_{4} = 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

## The solubility harmony throughout the saturated option would be AgCl (s) \(\rightleftharpoons\) Ag + (aq) + Cl – (aq) New solubility out-of AgCl was step one

- Suggest the difference between ionic product and you can solubility tool.
- The fresh new solubllity of AgCI within the water within 298 K was 1.06 x 10 -5 mole per litre. Estimate is actually solubility device at this temperature.

## Brand new solubility equilibrium in the saturated solution is AgCl (s) \(\rightleftharpoons\) Ag + (aq) + Cl – (aq) The latest solubility from AgCl try 1

- It’s applicable to types of selection.
- Their worthy of changes toward improvement in scam centration of ions.

## Brand new solubility equilibrium on saturated option would be AgCl (s) \(\rightleftharpoons\) Ag + (aq) + Cl – (aq) The fresh solubility out-of AgCl try step one

- It’s relevant toward over loaded selection.
- This has a particular really worth for an enthusiastic electrolyte on a constant 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: