In the event your solubility tool regarding direct iodide try 3

Question 1cuatro. dos x 10 -8 , its solubility will be ……….. Pearland escort… (a) 2 x 10 -3 M (b) 4 x 10 -4 M (c) l.6 x 10 -5 M (d) 1.8 x 10 -5 M Answer: (a) 2 x 10 -3 M PbI₂ (s) > Pb 2+ (aq) + 2I – (aq) K_sp = (s) (2s) 2 3.2 x 10 -8 = 4s 3

Concern 17

Question 15. ₂Y_(g) \(\rightleftharpoons\) 2X + + Y 2- (aq), calculate the solubility product of X₂Y in water at 300K (R = 8.3 J K -1 Mol -1 ) ………………. (a) 10 -10 (b) 10 -12 (c) 10 -14 (d) can not be calculated from the given data Answer: (a) 10 -10 KJ mol -1 = – 2.303 x 8.3 JK -1 mol -1 x 300K log K_sp

K_eq = [x + ] 2 [Y 2- ] ( X₂Y_(s) = 1) K_eq = K Question 16. MY and NY₃, are insoluble salts and have the same K_sp values of 6.2 x 10 -13 at room temperature. Which statement would be true with regard to MY and NY₃? (a) The salts MY and NY₃ are more soluble in O.5 M KY than in pure water (b) The addition of the salt of KY to the suspension of MY and NY₃ will have no effect on (c) The molar solubities of MY and NY₃ in water are identical (d) The molar solubility of MY in water is less than that of NY₃ Answer: (d) The molar solubility of MY in water is less than that of NY₃ Addition of salt KY (having a common ion Y – ) decreases the solubility of MY and NY₃ due to common ion effect. Option (a) and (b) are wrong. For salt MY, MY \(\rightleftharpoons\) M + + Y – K_sp = (s) (s) 6.2 x 10 -13 = s 2

What is the pH of your resulting services whenever equal quantities away from 0.1M NaOH and you can 0.01M HCl are mixed? (a) 2.0 (b) step three (c) 7.0 (d) Answer: (d) x ml out of 0.1 yards NaOH + x ml from 0.01 M HCI Zero. out of moles from NaOH = 0.step one x x x 10 -step three = 0.l x x ten -3 Zero. from moles from HCl = 0.01 x x x ten -3 = 0.01 x x ten -3 Zero. out of moles from NaOH once collection = 0.1x x ten -step three – 0.01x x ten -step three = 0.09x x ten -step 3 Intensity of NaOH =

[OH – ] = 0.045 p OH = – journal (4.5 x 10 -dos ) = 2 – record 4.5 = dos – 0.65 = step one.thirty-five pH = fourteen – step one.thirty five =

Question 18. The dissociation constant of a weak acid is 1 x 10 -3 . In order to prepare a buffer solution with a pH =4, the [Acid] / [Salt] ratio should be ……………….. (a) 4:3 (b) 3:4 (c) 10:1 (d) 1:10 Answer: (d) 1:10 K_a = 1 x 10 -3 ; pH = 4

Concern 19. The fresh pH regarding ten -5 Yards KOH provider might possibly be ………….. (a) nine (b) 5 (c)19 (d) not one of those Answer: (a) nine

[OH – ] = ten -5 Yards. pH = 14 – pOH . pH = 14 – ( – record [OH – ]) = fourteen + diary [OH – ] = 14 + diary 10 -5 = fourteen – 5 = nine

Playing with Gibb’s totally free opportunity change, ?G 0 = KJ mol -step 1 , toward response, X

Question 21. Which of the following can act as lowery – Bronsted acid well as base? (a) HCl (b) SO₄ 2- (c) HPO₄ 2- (d) Br – Answer: (c) HPO₄ 2- HPO₄ 2- can have the ability to accept a proton to form H₂PO₄. It can also have the ability to donate a proton to form PO₄ -3 .