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Thursday, September 29, 2016

Solubility equilibria of sparingly soluble salts

  1. The solubility depends on a number of factors important amongst which are the lattice enthalpy of the salt and the solvation enthalpy of the ions in a solution.
  2. For a salt to dissolve in a solvent the strong forces of attraction between its ions (lattice enthalpy) must be overcome by the ion-solvent interactions.
  3. The solvation enthalpy of ions is referred to in terms of solvation which is always negative i.e. energy is released in the process of solvation.
  4. The amount of solvation enthalpy depends on the nature of the solvent. In case of a non-polar (covalent) solvent, solvation enthalpy is small and hence, not sufficient to overcome lattice enthalpy of the salt. Consequently, the salt does not dissolve in non-polar solvent.
  5. As a general rule , for a salt to be able to dissolve in a particular solvent its solvation enthalpy must be greater than its lattice enthalpy so that the latter may be overcome by former. Each salt has its characteristic solubility which depends on temperature.

 Solubility product constant:

The solubility product constant, Ksp, is the equilibrium constant for a solid substance dissolving in an aqueous solution. It represents the level at which a solute dissolves in solution. The more soluble a substance is, the higher the Ksp value it has.

Consider the general dissolution reaction below:

aA (s) cC (aq) + dD (aq)

To solve for the Ksp it is necessary to take the molarities or concentrations of the products (cC and dD) and multiply them. If there are coefficients in front of any of the products, it is necessary to raise the product to that coefficient power (and also multiply the concentration by that coefficient). This is shown below:

Ksp = [C]c [D]d

Note that the reactant, aA, is not included in the Ksp equation. Solids are not included when calculating equilibrium constant expressions, because their concentrations do not change the expression; any change in their concentrations are insignificant, and therefore omitted.

Hence, Ksp represents the maximum amount of solid that can be dissolved in the aqueous solution.

The answer will have the units of molarity, mol L-1, a measure of concentration.

The solubility (by which we usually mean the molar solubility) of a solid is expressed as the concentration of the "dissolved solid" in a saturated solution. In the case of a simple 1:1 solid such as AgCl, this would just be the concentration of Ag+ or Cl in the saturated solution. But for a more complicated stoichiometry such as as silver chromate, the solubility would be only one-half of the Agconcentration.

For example, let us denote the solubility of Ag2CrO4 as S mol L–1. Then for a saturated solution, we have
Ag2CrO4(s) 2Ag++CrO24

[Ag+] = 2S    [CrO42–] = S.

Ksp = [Ag+]2 [CrO42–]

(2S)2 (S) = 4S3 = 2.76×10–12

S = (Ks/4)1/3 = (6.9 x 10-13)1/3 = (0.69 x 10-12)1/3 = 3√ (0.69) x 10-4 = 0.88 x 10-4

Thus the solubility is 8.8×10–5M.

Important effects:

  1. For highly soluble ionic compounds the ionic activities must be found instead of the concentrations that are found in slightly soluble solutions.
  2. Common Ion Effect: The solubility of the reaction is reduced by the common ion. For a given equilibrium, a reaction with a common ion present has a lower Ksp, and the reaction without the ion has a greater Ksp.
  3.  Salt Effect (diverse ion effect): Having an opposing effect on the Ksp value compared to the common ion effect, uncommon ions increase the Ksp value. Uncommon ions are ions other than those involved in equilibrium.
  4. Ion Pairs: With an ionic pair (a cation and an anion), the Ksp value calculated is less than the experimental value due to ions involved in pairing. To reach the calculated Ksp value, more solute must be added.
  5. Solubility S increases with increase in [H+] or decrease in pH.

Wednesday, September 28, 2016

Work Energy Theorem


The work-energy theorem is a generalized description of motion that states that the work done by the sum of all forces acting on an object is equal to the change in that object's kinetic energy.

f = ma

f = m dv/dt

We know that

dw = fdx

w = ∫ f. dx

= ∫ m. a dx

= ∫ m. (dv/dt) dx

= m ∫dv dx/dt

= m ∫ v. dv

Applying limits

Initial Velocity = u;

Final Velocity = v;

= m ∫uv v dv

= m [v²/2]uv = ½ mv² - ½ mu²

= Δ KE

So, Work done = Change in KE

Hence proved.

Tuesday, September 27, 2016

hyperbolic functions

  • ex = 1 + x/1! +x²/2! + x³/3! + … + xn / n! + … ∞

  • e-x = 1 - x/1! +x²/2! - x³/3! + … + (-1)n xn / n! + … ∞

  • sinhx = (ex – e-x) / 2 = x/1! +x²/3! + … + … ∞

  • coshx = (ex + e-x) / 2 = 1 +x²/2! + … ∞

  • tanhx = sinhx/ coshx = (ex – e-x)/ (ex + e-x)

  • sechx = 1/ coshx = 2/ (ex + e-x)

  • cosechx = 1/ sinhx = 2/ (ex + e-x)

  • cothx = 1/ tanhx = (ex + e-x)/ (ex – e-x)

  • sinh (-x) = - sinhx

  • cosh (-x) = coshx

  • tanh (-x) = - tanhx

  • sechx (-x) = sechx

  • cosech (-x) = - cosechx

  • sinh (x ± y) = sinhx coshy ± coshx sinhy

  • cosh (x ± y) = coshx coshy ± sinhx sinh

  • tanh (x ± y) = (tanhx ± tanhy)/ (1 ± tanhx tanhy)

  • sinh2x = 2 sinhx coshx = 2 tanhx/ (1 - tanh² x)

  • cosh2x = cosh²x + sinh²x = (1+ tanh²x)/ (1 - tanh²x)

  • tanh2x = 2tanhx/ (1 + tanh² x)

  • sinh2x + cosh2x = (1 + tanhx) / (1 – tanhx)

  • sinh3x = 3 sinhx + 4 sinh³x

  • cosh 3x = 4 cosh³x – 3 coshx

  • tanh 3x = (3 tanhx + tan³x)/ (1 + 3tanh²x)

  • sinh (x + y) sinh(x - y) = sinh³x - sinh²y

  • cosh (x + y) cosh (x - y) = cosh²x + sinh²y

  • (coshx + sinhx)n = (cosh[nx] + sinh [nx]) = enx

  • (coshx - sinhx)n = (cosh [nx] - sinh [nx]) = e-nx

  • cosh (2nx) + sinh (2nx) = [(1 + tanhx)/ (1 - tanhx)]n

[1, ∞)
(-1, 1)
R – {0}
R – [-1, 1]
R – {0}
R – {0}
(0, 1]

 Graphs of Hyperbolic functions
i) y = sinhx
ii) y = coshx
iii) y = tanhx
iv) y = cothx
v) y = sechx
vi) y = cosechx

Algebra Puzzle [With Answers] ......

Hello Everyone.....

Here are the answers of yesterday Algebra Puzzles......

Puzzle- 1:

Solve the following:


(Replace letters with digits and have the sum be true. A, B, C, D and E are all different digits.)

Answer: 2178 x 4 = 8172

Puzzle- 2:

Solve the following:


Answer: 437 + 589 = 1026


743 + 859 = 1602

Puzzle- 3:

Solve the following:


Answer: 285714 × 3 = 857142 


142857 × 3 = 428571

Puzzle- 4:

Solve the following:


Answer: 57 = 78125

Puzzle- 5:

Solve the following:

(L + O + G + I + C)3 = LOGIC

Answer: (1+9+6+8+3)3 = 19683

"Robbie" has worked out another solution (but it does use 7 twice)

(1+7+5+7+6)3 = 17576

Monday, September 26, 2016

Relationship between equilibrium constant K, reaction quotient Q and Gibbs energy G

We know that if,
  •  ΔG is negative, then the reaction is spontaneous and proceeds in the forward direction.
  • ΔG is positive, then reaction is considered non-spontaneous. Instead, as reverse reaction would take place
  •  ΔG is 0, reaction has achieved equilibrium; at this point, there is no longer any free energy left to drive the reaction.

A mathematical expression of this thermodynamic view of equilibrium can be described by the following equation:
ΔG = ΔG⁰ + RT ln Q

Where, G is standard Gibbs energy.

At equilibrium, when ΔG = 0 and Q = Kc

ΔG = ΔG⁰ + RT ln K = 0

ΔG⁰ = -RT ln K

ln K = ΔG⁰ / RT
Taking antilog of both sides, we get,
K = e-ΔG⁰ /RT
  • If ΔG0 < 0, then –ΔG0/RT is positive, and e-ΔG⁰ /RT >1, making K >1, which implies a spontaneous reaction or the reaction which proceeds in the forward direction to such an extent that the products are present predominantly.
  • If ΔG0 > 0, then –ΔG0/RT is negative, and e-ΔG⁰ /RT < 1, that is, K < 1, which implies a non-spontaneous reaction or a reaction which proceeds in the forward direction to such a small degree that only a very minute quantity of product is formed.

Algebra Puzzle.........

Hello Everyone.....

Here are some Algebra Puzzles give a try....

Puzzle- 1:

Solve the following:


(Replace letters with digits and have the sum be true. A, B, C, D and E are all different digits.)

Puzzle- 2:

Solve the following:


Puzzle- 3:

Solve the following:


Puzzle- 4:

Solve the following:


Puzzle- 5:

Solve the following:

(L + O + G + I + C)3 = LOGIC

Sunday, September 25, 2016

Weightlessness in Satellites:

           The weight of a body is felt due to a reactionary force applied on the body by some other body (which is in contact with the first body). For example, when we stand on a plane, we feel our weight due to the reaction of the plane on our feet. If under some special circumstances the reaction of this plane becomes zero then we shall feel as our weight has also become zero. This is called "state of weightlessness". If the ropes of a descending lift are broken, then persons standing in the lift will feel this state. Weightlessness is also felt by a space-man inside an artificial satellite. Suppose an artificial satellite of mass m is revoking around the earth (mass Me) with speed v0 in an orbit of radius r. The necessary centripetal force is provided by the gravitational force:

The mass of a person doesn’t change but their apparent weight does change when accelerating or in a free fall.

GMem/r² = mv²₀/2
GMe/r² = v²₀/2 … (i)

If there is a space-man of mass m’ inside the satellite, he is acted upon by two forces:

  • Gravitational force GMem/r²,
  • Reaction R of the base of the satellite, in the opposite direction. Thus, there is a net force [GMem’/r² - R] on the man.
It is directed towards the centre of the orbit and is the necessary centripetal force on the man. That is,

[GMem’/r² - R]  = m’v²₀ / r
GMe/r² - R/m’ = v²₀ / 2

Substituting the value of v²₀ / r from eq. (1), we get

GMe/r² - R/m’ = GMe/r²

.˙. R = 0

Thus, the reactionary force on the man is zero. Hence he feels his weight zero. If he stands on a spring-balance, the balance will read zero. In fact, everybody inside the satellite is in a state of weightlessness.