If your solubility device out of lead iodide is actually step three

If your solubility device 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) 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 chicas escort League City TX 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 ensuing provider when equivalent volumes of 0.1M NaOH and you may 0.01M HCl was mixed? (a) dos.0 (b) step three (c) eight.0 (d) Answer: (d) x ml away from 0.step 1 meters NaOH + x ml from 0.01 Yards HCI No. of moles from NaOH = 0.step 1 x x x ten -step 3 = 0.l x x 10 -step three Zero. regarding moles of HCl = 0.01 x x x 10 -3 = 0.01 x x 10 -3 No. regarding moles regarding NaOH just after combo = 0.1x x 10 -step three – 0.01x x 10 -step 3 = 0.09x x ten -step three Intensity of NaOH =

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