When your solubility device off lead iodide is step step step step step step step 3

When your solubility device off lead iodide is step step step step step step step 3

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) Ksp = (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 Ksp

Keq = [x + ] 2 [Y 2- ] ( X2Y(s) = 1) Keq = K Question 16. MY and NY3, are insoluble salts and have the same Ksp 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 – Ksp = (s) (s) 6.2 x Jackson escort service 10 -13 = s 2

What’s the pH of ensuing solution whenever equivalent amounts out of 0.1M NaOH and you may 0.01M HCl is mixed? (a) dos.0 (b) step 3 (c) seven.0 (d) Answer: (d) x ml from 0.step one m NaOH + x ml away from 0.01 Yards HCI No. out-of moles regarding NaOH = 0.step one x x x 10 -step three = 0.l x x 10 -step three Zero. out-of moles regarding HCl = 0.01 x x x 10 -3 = 0.01 x x ten -3 Zero. regarding moles off NaOH after blend = 0.1x x ten -step three – 0.01x x ten -step 3 = 0.09x x 10 -step three Concentration of NaOH =

[OH – ] = 0.045 p OH = – log (4.5 x ten -2 ) = 2 – journal 4.5 = dos – 0.65 = 1.thirty five pH = 14 – step 1.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 Ka = 1 x 10 -3 ; pH = 4

Question 19. The newest pH away from ten -5 Yards KOH provider might possibly be ………….. (a) 9 (b) 5 (c)19 (d) none ones Respond to: (a) nine

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

Playing with Gibb’s free opportunity alter, ?G 0 = KJ mol -1 , with 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 .

Leave a Comment

Your email address will not be published. Required fields are marked *

Shopping Cart