When your solubility equipment out-of lead iodide is actually 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 escort services in Simi Valley + + Y – K_sp = (s) (s) 6.2 x 10 -13 = s 2

What is the pH of ensuing services when equal volumes out of 0.1M NaOH and you can 0.01M HCl is actually combined? (a) dos.0 (b) step three (c) seven.0 (d) Answer: (d) x ml out-of 0.1 meters NaOH + x ml out of 0.01 Yards HCI Zero. from moles out-of NaOH = 0.1 x x x ten -step three = 0.l x x 10 -step three Zero. of moles out-of HCl = 0.01 x x x ten -step three = 0.01 x x 10 -step 3 No. off moles out-of NaOH immediately following mix = 0.1x x ten -step three – 0.01x x ten -step 3 = 0.09x x 10 -step three Intensity of NaOH =

[OH – ] = 0.045 p OH = – journal (cuatro.5 x 10 -dos ) = 2 – log cuatro.5 = 2 – 0.65 = 1.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. This new pH of 10 -5 Meters KOH service was ………….. (a) 9 (b) 5 (c)19 (d) nothing of those Respond to: (a) nine

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

Playing with Gibb’s 100 % free energy transform, ?Grams 0 = KJ mol -step 1 , to the effect, 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 .