If your solubility equipment out of head iodide try step step step step step step three

If your solubility equipment out of head iodide try step step step step step 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) 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 the ensuing provider when equivalent amounts from 0.1M NaOH and you will 0.01M HCl is actually combined? (a) dos.0 (b) step 3 (c) seven.0 (d) Answer: (d) x ml regarding 0.step Sparks escort one yards NaOH + x ml away from 0.01 Yards HCI No. from moles out of NaOH = 0.step 1 x x x 10 -step three = 0.l x x ten -3 Zero. away from moles from HCl = 0.01 x x x ten -step 3 = 0.01 x x 10 -step 3 Zero. from moles off NaOH after mixing = 0.1x x ten -step 3 – 0.01x x 10 -step 3 = 0.09x x 10 -step 3 Concentration of NaOH =

[OH – ] = 0.045 p OH = – diary (cuatro.5 x ten -2 ) = 2 – log 4.5 = 2 – 0.65 = step 1.thirty five pH = 14 – 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 Ka = 1 x 10 -3 ; pH = 4

Matter 19. The pH away from 10 -5 Meters KOH service was ………….. (a) 9 (b) 5 (c)19 (d) none of them Respond to: (a) nine

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

Having fun with Gibb’s totally free times alter, ?Grams 0 = KJ mol -1 , on impulse, 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 .

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