The latest pH regarding a sample regarding white vinegar is actually step three

Question 13. 76. Calculate the concentration of hydrogen ion in it. Answer: pH = – log_ten [H₃O + ] = – log₁₀ = – pH = – 3.76 = \(\overline<4>\).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 base is F K_b = K_w/K_a = \(\frac<1>><6.8>>\) = l.47 x 10 -eleven = l.5 x 10 -11

Matter 15. This new pH from 0.step one Meters provider regarding cyanic acidic (HCNO) is 2.34. Calculate the fresh new ionization ongoing of the acid and its own level of ionization in the solution. HCNO \(\rightleftharpoons\) H + + CNO – pH = 2.34 mode – log [H + ] = 2.34 or journal [H + ] female escort Oxnard CA = – dos.34 = step 3.86 otherwise [H + ] = Antilog step 3.86 = cuatro.57 x ten -step 3 Meters [CNO – ] = [H + ] = cuatro.57 x 10 -step three 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 10 -7 pOH = – record (9.44 x 10 -seven ) = seven – 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: 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

The latest solubility balance throughout the soaked solution is AgCl (s) \(\rightleftharpoons\) Ag + (aq) + Cl – (aq) The fresh solubility off AgCl is actually 1

1. Point out the distinctions between ionic device and you will solubility tool.
2. The solubllity away from AgCI in the water during the 298 K are step 1.06 x 10 -5 mole for every single litre. Assess are solubility tool at that temperature.

The new solubility equilibrium in the over loaded solution is AgCl (s) \(\rightleftharpoons\) Ag + (aq) + Cl – (aq) Brand new solubility off AgCl is actually 1

1. It is appropriate to any or all brand of alternatives.
2. The value change on change in ripoff centration of ions.

The fresh solubility harmony on the saturated solution is AgCl (s) \(\rightleftharpoons\) Ag + (aq) + Cl – (aq) The solubility regarding AgCl is step 1

1. It is relevant to the soaked choices.
2. It offers a definite worthy of to have a keen electrolyte within 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: