## In can excellentse your solubility tool away from direct iodide are 3

In can excellentse your solubility tool away from direct iodide are 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 10 -13 = s 2

What is the pH of your own ensuing service when equivalent volumes regarding 0.1M NaOH and you will 0.01M HCl try blended? (a) 2.0 (b) 3 (c) eight.0 (d) Answer: (d) x ml out-of 0.1 yards NaOH + x ml regarding 0.01 Meters HCI Zero. out of moles regarding NaOH = 0.1 x x x 10 -step three = 0.l x x ten -step three No. off moles out-of HCl = 0.01 x x x ten -3 = 0.01 x x ten -step three Zero. regarding moles out of NaOH shortly after collection = 0.1x x 10 -step three – 0.01x x 10 -step 3 = 0.09x x ten -step 3 Intensity of NaOH =