Whether your solubility product out of lead iodide was step three

Question 1cuatro. dos 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

Question 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 ensuing service when equal amounts regarding 0.1M NaOH and you may 0.01M HCl are combined? (a) 2.0 (b) step 3 (c) seven.0 (d) Answer: (d) x ml from 0.step one meters NaOH + x ml away from 0 babylon escort Riverside.01 M HCI No. off moles of NaOH = 0.step one x x x ten -3 = 0.l x x ten -step three Zero. regarding moles of HCl = 0.01 x x x ten -step three = 0.01 x x ten -step three No. away from moles of NaOH after blend = 0.1x x 10 -step three – 0.01x x 10 -step three = 0.09x x 10 -step 3 Intensity of NaOH =

[OH – ] = 0.045 p OH = – journal (4.5 x 10 -2 ) = 2 – record 4.5 = dos – 0.65 = step one.thirty-five pH = 14 – step one.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

Question 19. Brand new pH out of ten -5 Meters KOH services is ………….. (a) 9 (b) 5 (c)19 (d) nothing of those Address: (a) 9

[OH – ] = 10 -5 Yards. pH = fourteen – pOH . pH = fourteen – ( – log [OH – ]) = 14 + log [OH – ] = 14 + diary ten -5 = fourteen – 5 = 9

Using Gibb’s 100 % free opportunity transform, ?G 0 = KJ mol -step 1 , towards reaction, 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 .