Question 13. 76. Calculate the concentration of hydrogen ion in it. Answer: pH = – log10 [H3O + ] = – log10 = – pH = – 3.76 = \(\overline
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
- Mention the differences anywhere between ionic equipment and solubility tool.
- 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
- It’s relevant to all or any particular alternatives.
- 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
- It’s applicable into the saturated choice.
- 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: