The latest pH off an example out of white vinegar is actually step step three
Question 13. 76. Calculate the concentration of hydrogen ion in it. Answer: pH = – logten [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 -eleven = l.5 x 10 -11
Question fifteen. Brand new pH out-of 0.step 1 M services out of cyanic acid (HCNO) is 2.34. Estimate the newest ionization ongoing of acid and its standard of ionization in the services. HCNO \(\rightleftharpoons\) H + + CNO – pH = 2.34 setting – log [H + ] = 2.34 or diary [H + ] = – 2.34 = 3.86 or [H + ] = Antilog 3.86 = 4.57 x ten -step 3 M [CNO – ] = [H + ] = cuatro.57 x 10 -step three Yards
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 ten -eight pOH = – journal (nine.49 x 10 -7 ) = eight – 0.9750 = six.03 pH = fourteen – pOH = 14 – six.03 = seven.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 equilibrium on the saturated option would be AgCl (s) \(\rightleftharpoons\) Ag + (aq) + Cl – (aq) New solubility out-of AgCl is actually step one
- Point out the differences between ionic equipment and solubility equipment.
- The brand new solubllity from AgCI in water at 298 K try step 1.06 x ten -5 mole each litre. Determine is actually solubility product at this temperature.
The newest solubility balance in the soaked solution is AgCl (s) \(\rightleftharpoons\) Ag + (aq) + Cl – (aq) New solubility out-of AgCl are 1
- It is applicable to all or any version of choice.
- The really worth changes to the change in swindle centration of one’s ions.
The latest solubility equilibrium on over loaded solution is AgCl (s) \(\rightleftharpoons\) Ag + (aq) + Cl – (aq) The brand new solubility out of AgCl are step one
- It is appropriate on the over loaded choices.
- This has a definite really worth for an enthusiastic electrolyte on a stable 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 Chula Vista backpage female escort are 2.0 x 10 -15 and 6 x 10 -17 respectively. Which salt is more soluble? Explain. Answer: