The latest pH out-of a sample off vinegar is step three

Question 13. 76. Calculate the concentration of hydrogen ion in it. Answer: pH = – log10 [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 Kb = Kw/Ka = \(\frac<1>><6.8>>\) = l.47 x 10 -11 = l.5 x 10 -11

Concern 15. The fresh new pH away from 0.step 1 Meters solution from cyanic acid (HCNO) is dos.34. Determine new ionization ongoing of your own acidic and its particular level of ionization throughout the service. HCNO \(\rightleftharpoons\) H + + CNO – pH = 2.34 means – journal [H + ] = dos.34 or log [H + ] = – 2.34 = step 3.86 or [H + ] = Antilog step three.86 = 4.57 x 10 -step three Yards [CNO – ] = [H + ] = 4.57 x ten -step three M

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 dos.thirty six x ten -5 = 944 x 10 -eight pOH = – diary (nine.49 x ten -seven ) = 7 – 0.9750 = 6.03 pH = 14 – pOH = fourteen – 6.03 = 7.97

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

This new solubility equilibrium throughout the soaked solution is AgCl (s) \(\rightleftharpoons\) Ag + (aq) + Cl – (aq) The latest solubility off AgCl try 1

  1. Mention the differences anywhere between ionic equipment and solubility tool.
  2. The brand new solubllity off AgCI within the water at the 298 K try step 1.06 x ten -5 mole for each litre. Assess is solubility product at that temperatures.

New solubility harmony from the soaked option would be AgCl (s) \(\rightleftharpoons\) Ag + (aq) + Cl – (aq) Brand new solubility from AgCl is actually step one

  1. It’s relevant to all or any particular alternatives.
  2. Their well worth alter into change in ripoff centration of the ions.

The fresh solubility equilibrium regarding soaked option would be AgCl (s) \(\rightleftharpoons\) Ag + (aq) + Cl – (aq) The fresh solubility away from AgCl try step 1

  1. It’s applicable into the saturated choice.
  2. This has a definite well worth to possess a keen electrolyte within a stable temperatures.

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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: