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
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 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 10 -13 = s 2
What is the pH of your ensuing provider when equivalent amounts away from 0.1M NaOH and you will 0.01M HCl are combined? (a) 2.0 (b) step three (c) eight.0 (d) Answer: (d) x ml from 0.step one m NaOH + x ml from Pueblo CO escort review 0.01 M HCI No. off moles off NaOH = 0.step 1 x x x 10 -3 = 0.l x x ten -3 Zero. out of moles out of HCl = 0.01 x x x ten -3 = 0.01 x x ten -step three Zero. away from moles out of NaOH after mixing = 0.1x x ten -3 – 0.01x x 10 -step three = 0.09x x 10 -3 Intensity of NaOH =
[OH – ] = 0.045 p OH = – journal (4.5 x 10 -dos ) = 2 – log cuatro.5 = dos – 0.65 = 1.thirty-five pH = fourteen – 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
Matter 19. Brand new pH from ten -5 Meters KOH services might possibly be ………….. (a) nine (b) 5 (c)19 (d) none of them Address: (a) 9
[OH – ] = ten -5 Meters. pH = fourteen – pOH . pH = 14 – ( – log [OH – ]) = 14 + journal [OH – ] = fourteen + log ten -5 = 14 – 5 = nine
Using Gibb’s totally free energy transform, ?Grams 0 = KJ mol -step one , towards the 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 .