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The bra beneficialnd new pH out of a sample away from vinegar is actually step three

utworzone przez | cze 27, 2022 | olathe escort | 0 komentarzy

The bra beneficialnd new pH out of a sample away from vinegar is actually step three

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, 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

Concern fifteen. The fresh pH regarding 0.1 Meters solution of cyanic acid (HCNO) is actually 2.34. Determine the ionization constant of your acid and its degree of ionization about provider. HCNO \(\rightleftharpoons\) H + + CNO – pH = dos.34 means – log [H + ] = 2.34 or record [H + ] = – 2.34 = 3.86 otherwise [H + ] = Antilog 3.86 = 4.57 x ten -step three M [CNO – ] = [H + ] = cuatro.57 x ten -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.36 x ten -5 = 944 x 10 -seven pOH = – record (9.49 x 10 -7 ) = seven – 0.9750 = six.03 pH = 14 – pOH = 14 – six.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

The new solubility equilibrium throughout the soaked option would be AgCl (s) \(\rightleftharpoons\) escort reviews Olathe Ag + (aq) + Cl – (aq) The new solubility of AgCl is 1

  1. Highlight the difference ranging from ionic product and you will solubility product.
  2. New solubllity off AgCI in the water during the 298 K is actually step 1.06 x ten -5 mole each litre. Assess was solubility unit at that temperature.

The latest solubility harmony regarding over loaded solution is AgCl (s) \(\rightleftharpoons\) Ag + (aq) + Cl – (aq) The new solubility away from AgCl try step one

  1. It is appropriate to all kind of choices.
  2. Its value transform to your change in ripoff centration of the ions.

The fresh solubility balance in the over loaded option would be AgCl (s) \(\rightleftharpoons\) Ag + (aq) + Cl – (aq) The fresh new solubility off AgCl are step 1

  1. It’s appropriate towards the saturated alternatives.
  2. It offers one particular well worth for an electrolyte in the a stable 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 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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