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The fresh pH away from an example off vinegar is step three

The fresh pH away from an example off vinegar is step three

Question 13. 76. Calculate the concentration of hydrogen ion in it. Answer: pH = – logten [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 -eleven = l.5 x 10 -11

Matter 15. This new pH out-of 0.1 Meters service off cyanic acid (HCNO) try dos.34. Determine the ionization constant of your own acidic as well as degree of ionization in the solution. HCNO \(\rightleftharpoons\) H + + CNO – pH = 2.34 setting – journal [H + ] = 2.34 or journal [H + ] = – dos.34 = step three.86 otherwise [H + ] = Antilog 3.86 = cuatro.57 x 10 -step three Yards [CNO – ] = [H + ] = 4.57 x ten -step 3 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 ten -7 pOH = – journal (nine.forty-two x 10 -eight ) = 7 – 0.9750 = 6.03 pH = fourteen – pOH = 14 – 6.03 Jacksonville escort service = 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

The latest solubility harmony on the over loaded solution is AgCl (s) \(\rightleftharpoons\) Ag + (aq) + Cl – (aq) New solubility off AgCl are 1

  1. Suggest the differences ranging from ionic product and you can solubility equipment.
  2. The latest solubllity regarding AgCI in the water in the 298 K was 1.06 x 10 -5 mole for each litre. Calculate was solubility unit at that heat.

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

  1. It’s appropriate to types of alternatives.
  2. Its well worth changes for the improvement in swindle centration of one’s ions.

The solubility harmony on saturated solution is AgCl (s) \(\rightleftharpoons\) Ag + (aq) + Cl – (aq) This new solubility out of AgCl is actually 1

  1. It is appropriate towards saturated solutions.
  2. It’s got a particular really worth to own an electrolyte within a steady 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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