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The new pH away from a sample away from white vinegar is 3

The new pH away from a sample away from white vinegar is 3

Question 13. 76. Calculate the concentration of hydrogen ion in it. Answer: pH = – logten [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 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 Kb = Kw/Ka = \(\frac<1>><6.8>>\) = 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, Kh = 2.22 x 10-11 Kw/Kb = 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, Ksp = 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 Ksp = [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 solubility harmony throughout the saturated option would be AgCl (s) \(\rightleftharpoons\) Ag + (aq) + Cl – (aq) New solubility out-of AgCl was step one

  1. Suggest the difference between ionic product and you can solubility tool.
  2. 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

  1. It’s applicable to types of selection.
  2. 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

  1. It’s relevant toward over loaded selection.
  2. 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 Ksp = [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:

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