The fresh new pH out-of a sample out of white vinegar is 3

24 Juin The fresh new pH out-of a sample out of white vinegar is 3

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 -eleven = l.5 x 10 -11

Matter fifteen. New pH off 0.step 1 Meters service off cyanic acid (HCNO) is dos.34. Determine the fresh ionization constant of acid and its particular level of ionization regarding the service. HCNO \(\rightleftharpoons\) H + + CNO – pH = dos.34 mode – record [H + ] = 2.34 or journal [H + ] = – 2.34 = 3.86 otherwise [H + ] = Antilog step three.86 = 4.57 x 10 -3 M [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 dos.thirty-six x 10 -5 = 944 x 10 -seven pOH = – record (9.forty two x 10 -seven ) = eight – 0.9750 = six.03 pH = fourteen – pOH = 14 – six.03 = 7.97

Question 17. What is the minimum volume of water required to dissolve 1 eros escort High Point 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

Brand new solubility equilibrium throughout the saturated solution is AgCl (s) \(\rightleftharpoons\) Ag + (aq) + Cl – (aq) The new solubility away from AgCl is actually 1

1. Mention the differences between ionic product and solubility unit.
2. The fresh new solubllity out of AgCI in the water within 298 K was step one.06 x 10 -5 mole each litre. Calculate is actually solubility device at that heat.

The brand new solubility harmony on soaked solution is AgCl (s) \(\rightleftharpoons\) Ag + (aq) + Cl – (aq) The brand new solubility of AgCl are 1

1. It is relevant to all type of solutions.
2. The worthy of transform on change in swindle centration of the ions.

The fresh new solubility equilibrium regarding the soaked option would be AgCl (s) \(\rightleftharpoons\) Ag + (aq) + Cl – (aq) The brand new solubility away from AgCl are 1

1. It is applicable toward soaked options.
2. It has got a particular well worth to own an electrolyte from the a constant heat.

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: