The latest pH of a sample from vinegar is step 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

Concern fifteen. Brand new pH from 0.step 1 M services from cyanic acidic (HCNO) is actually dos.34. Determine the new ionization ongoing of your acid and its level of ionization from the solution. HCNO \(\rightleftharpoons\) H + + CNO – pH = dos.34 setting – journal [H + ] = dos.34 or record [H + ] = – 2.34 = step three.86 otherwise [H + ] = Antilog step three.86 = 4.57 x 10 -step 3 Yards [CNO – ] = [H + ] = cuatro.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 2.thirty six x 10 -5 = 944 x ten -7 pOH = – log (9.forty-two x ten -eight ) = 7 – 0.9750 = 6.03 pH = fourteen – pOH = 14 – 6.03 = seven.97

Question 17. What is the minimum https://datingranking.net/escort-directory/meridian/ 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 balance from the soaked option would be AgCl (s) \(\rightleftharpoons\) Ag + (aq) + Cl – (aq) The latest solubility regarding AgCl was 1

  1. Suggest the difference ranging from ionic tool and you can solubility equipment.
  2. The solubllity away from AgCI within the water from the 298 K is step 1.06 x 10 -5 mole for each litre. Determine try solubility tool at that heat.

The newest solubility harmony regarding soaked option would be AgCl (s) \(\rightleftharpoons\) Ag + (aq) + Cl – (aq) The newest solubility out-of AgCl is 1

  1. It is relevant to all form of choice.
  2. The well worth alter for the change in scam centration of one’s ions.

The latest solubility balance regarding the saturated option would be AgCl (s) \(\rightleftharpoons\) Ag + (aq) + Cl – (aq) This new solubility out-of AgCl was 1

  1. It’s relevant to the over loaded solutions.
  2. It offers one well worth for a keen electrolyte in the a steady 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 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: