If your solubility product off direct iodide are step step step 3

Question 1cuatro. 2 x 10 -8 , its solubility will be ………….. (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

Matter 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 the resulting provider when equal quantities away from 0.1M NaOH and you may babylon escort Salinas CA 0.01M HCl try mixed? (a) dos.0 (b) step three (c) seven.0 (d) Answer: (d) x ml off 0.1 yards NaOH + x ml of 0.01 M HCI No. out of moles away from NaOH = 0.step one x x x 10 -step three = 0.l x x ten -step 3 No. off moles off HCl = 0.01 x x x 10 -3 = 0.01 x x ten -3 No. away from moles regarding NaOH just after blend = 0.1x x 10 -3 – 0.01x x 10 -step three = 0.09x x 10 -step 3 Concentration of NaOH =

[OH – ] = 0.045 p OH = – record (4.5 x 10 -dos ) = dos – journal cuatro.5 = dos – 0.65 = step one.thirty-five pH = 14 – 1.35 =

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. Brand new pH away from 10 -5 Meters KOH provider is ………….. (a) 9 (b) 5 (c)19 (d) nothing of them Answer: (a) 9

[OH – ] = ten -5 Yards. pH = fourteen – pOH . pH = 14 – ( – journal [OH – ]) = fourteen + record [OH – ] = 14 + log ten -5 = 14 – 5 = 9

Having fun with Gibb’s totally free times transform, ?Grams 0 = KJ mol -step 1 , to your 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 .