Chemistry, asked by Psatyam7397, 11 months ago

Solubility product of barium sulphate at 25C is 1×10^-9 .What would be the conc of sulphuric acid necessary to ppt barium sulphate from the solution .If the sol contain 0.1 M barium ions

Answers

Answered by Sandeshjpillai

Answer:

Ksp = [Al3+] [OH–]3 = (S) (3S)3 = 27 S4

S4 = Ksp/27 = 27 × 10–11/27 x 10 = 1 × 10–12

S = 1× 10–3 mol L–1

(i) Solubility of Al(OH)3

Molar mass of Al (OH)3 is 78 g. Therefore,

Solubility of Al (OH)3 in g L–1 = 1 × 10–3 × 78 g L–1 = 78 × 10–3 g L–1 = 7.8 × 10–2 g L–1

(ii) pH of the solution

S = 1×10–3 mol L–1

[OH– ] = 3S = 3×1×10–3 = 3 × 10–3

pOH = 3 – log 3

pH = 14 – pOH = 11 + log 3 = 11.4771

35. Ksp of PbCl2 = 3.2 × 10–8

Let S be the solubility of PbCl2.

Ksp = [Pb2+] [Cl–]2 = (S) (2S)2 = 4S3

Ksp = 4S3

S3 = Ksp/4 = 3.2 x 10–8/4 mol L–1 = 8 × 10–9 mol L–1

Molar mass of PbCl2 = 278

∴ Solubility of PbCl2 in g L–1= 2 × 10–3 × 278 g L–1

= 556 × 10–3 g L–1

= 0.556 g L–1

To get saturated solution, 0.556 g of PbCl2 is dissolved in 1 L water.

0.1 g PbCl2 is dissolved in 0 .1/0 .556 L = 0.1798 L water.

To make a saturated solution, dissolution of 0.1 g PbCl2 in 0.1798 L ≈ 0.2 L of water will be required.

37. ΔrHΘ = ΔfHΘ[CaO(s)] + ΔfHΘ [CO2(g)] – ΔfHΘ [CaCO3(s)]

∴ ΔrHΘ = 178.3 kJ mol–1

The reaction is endothermic. Hence, according to Le-Chatelier’s principle, reaction will proceed in forward direction on increasing temperature.

IV. Matching Type

38. (i) → (b) (ii) → (d) (iii) → (c) (iv) → (a)

39. (i) → (d) (ii) → (c) (iii) → (b)

40. (i) → (d) (ii) → (a) (iii) → (b)

41. (i) → (b) (ii) → (e) (iii) → (c) (iv) → (d)

42. (i) → (c) (ii) → (a) (iii) → (b)

43. (i) → (b) and (c) (ii) → (d) (iii) → (a)

V. Assertion and Reason Type

44. (i) 45. (i) 46. (ii) 47.(iii) 48. (i) 49. (iii) 50. (iv)

VI. Long Answer Type

51. (i) Qc < Kc

(ii) Qc > Kc

(iii) Qc = Kc

where, Qc is reaction quotient in terms of concentration and Kc is equilibrium constant.

53.

S moles of AxBy dissolve to give xS moles of Ap+ and y S moles of Bq–.]

54. ΔG = ΔGΘ + RTlnQ

ΔGΘ = Change in free energy as the reaction proceeds

ΔG = Standard free energy change

Q = Reaction quotient

R = Gas constant

T = Absolute temperature

Since ΔGΘ = – RT lnK

∴ ΔG = – RT lnK + RT lnQ = RT ln Q/K

If Q < K, ΔG will be negative. Reaction proceeds in the forward direction.

If Q = K, ΔG = 0, no net reaction.

[Hint: Next relate Q with concentration of CO, H2, CH4 and H2O in view of reduced volume (increased pressure). Show that Q < K and hence the reaction proceeds in forward direction.]

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