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 Ag⁺ concentration.

For example, let us denote the solubility of Ag₂CrO₄ as S mol L^–1. Then for a saturated solution, we have

Ag₂CrO₄(s) ⇌ 2Ag⁺+CrO₂^–4

[Ag⁺] = 2S    [CrO₄^2–] = S.

Ksp = [Ag+]² [CrO₄^2–]

(2S)² (S) = 4S³ = 2.76×10^–12

S = (K_s/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.

Work Energy Theorem

Principle:

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 ∫_u^v v dv

= m [v²/2]_u^v = ½ mv² - ½ mu²

= Δ KE

So, Work done = Change in KE

Hence proved.

hyperbolic functions

• e^x = 1 + x/1! +x²/2! + x³/3! + … + xⁿ / n! + … ∞
  
  
• e^-x = 1 - x/1! +x²/2! - x³/3! + … + (-1)ⁿ xⁿ / n! + … ∞
  
  
• sinhx = (e^x – e^-x) / 2 = x/1! +x²/3! + … + … ∞
  
  
• coshx = (e^x + e^-x) / 2 = 1 +x²/2! + … ∞
  
  
• tanhx = sinhx/ coshx = (e^x – e^-x)/ (e^x + e^-x)
  
  
• sechx = 1/ coshx = 2/ (e^x + e^-x)
  
  
• cosechx = 1/ sinhx = 2/ (e^x + e^-x)
  
  
• cothx = 1/ tanhx = (e^x + e^-x)/ (e^x – 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]) = e^nx
  
  
• (coshx - sinhx)ⁿ = (cosh [nx] - sinh [nx]) = e^-nx
  
  
• cosh (2nx) + sinh (2nx) = [(1 + tanhx)/ (1 - tanhx)]ⁿ

Function	Domain	Range
sinhx	R	R
coshx	R	[1, ∞)
tanhx	R	(-1, 1)
cothx	R – {0}	R – [-1, 1]
cosechx	R – {0}	R – {0}
sechx	R	(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:

ABCD × E = DCBA

(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:

ABC + DEF = GHIJ

Answer: 437 + 589 = 1026

or

743 + 859 = 1602

Puzzle- 3:

Solve the following:

ABCDEF × 3 = BCDEFA

Answer: 285714 × 3 = 857142 

or 

142857 × 3 = 428571

Puzzle- 4:

Solve the following:

C^L = LOGIC

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

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^⁰ is standard Gibbs energy.

At equilibrium, when ΔG = 0 and Q = K_c

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:

ABCD × E = DCBA

(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:

ABC + DEF = GHIJ

Puzzle- 3:

Solve the following:

ABCDEF × 3 = BCDEFA

Puzzle- 4:

Solve the following:

C^L = LOGIC

Puzzle- 5:

Solve the following:

(L + O + G + I + C)³ = LOGIC