#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-5470-145

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

Question: 1) The Ksp of Ag2CrO4 is 1.1 ×10–12 at 298K. The solubility (in mol/L) of Ag2CrO4 in a 0.1M AgNO3 solution is   (IIT-JEE- 2013)

1) 1.1×10-11
2) 1.1×10-10
3) 1.1×10-9
4) 1.1×10-12

Solution:

Ag2+ is a common ion here. Hence, the contribution of Ag+ from AgNO3 would be ignored, and the solubility of Ag2CrO4 will depend on CrO4-2 only.

Let the concentration of CrO4-2 in solution = x molL-1

Ksp = 1.1 ´10–12 = [Ag+][CrO4-2]

or

1.1´10–12 = [0.1]2[x]

or

X = 1.1×10-10

Hence, the correct option is B. Question 2) Solubility product constant (KSP) of salts of type MX, MX2 and M3X at temperature “T” are 4.0×10-8, 3.2×10-14 and 2.7×10-15, respectively. Solubilities (mol dm-3) of the salts at temperature ‘T’ are    (IIT JEE -2008)

1) MX> MX2 > M3X
2) M3X> MX2> MX
3) MX2>M3X> MX
4) MX  >M3X> MX2

Solution:

For salt MX:

MX → M+ +X-

Ksp = [M+][X-]

Here [M+] = [X-]

So, Ksp =[M+]2

or

[M+] =[X-] =√Ksp = √(4.0×10-8) = 2×10-4

For salt MX2:

MX2 → M+ +2X-

Ksp = [M+][X-]2

Here 2[M+] = [X-]

So, Ksp =4[M+]3

or

[M+] =?Ksp/4 = ? [(3.2×10-14 )/4] = 2×10-5

For salt M3X,

M3X → 3M+ +X-

Ksp = [M+]3[X-]

Here [M+] = 3[X-]

So, Ksp =27[X-]4

or

[X-]4 = Ksp/27 =(2.7×10-15)/27

or

[X-] = 10-4

So, order of the solubilies is

MX  >M3X> MX2

Hence, the correct option is D.

Question: 3) In the following equilibrium, N2O4(g)↔ NO2(g) When 5 moles of each are taken, the temperature is kept at 298 K the total pressure was found to be 20 bar. Given that

ΔG0f ( N2O4) = 100 KJ

ΔG0f ( NO2) = 50 KJ

(i)     Find ΔG of the reaction

(ii)   The direction of the reaction in which the equilibrium shifts.   (IIT IEE -2004)

Solution:

(i) Given that,  P total = P (N2O4) = P (NO2) = 20 bar

But at equilibrium, P (N2O4) = P (NO2) ,

So, P (N2O4) = P (NO2) = 10 bar

Now

For the equilibrium reaction,

N2O4(g)↔ 2NO2(g)

Q = [P (NO2)]2 /[P (N2O4)] = 100/10 =

Now,

ΔG0 = 2 ΔG0f ( NO2) - ΔG0f ( N2O4) = 2×50 -100 = 0

So,

ΔG = ΔG0 – 2.303RT log Qp = 0 – 2.303×8.314×298×log 10

= -5706 J = -5.706 kJ

(ii) Negative value of ΔG indicates that reaction is spontaneous one and it will shift in forward direction.