Home
Class 12
CHEMISTRY
A very dilute saturated solution of a sp...

A very dilute saturated solution of a sparingly soluble salt `A_(3)B_(4)` has a vapour pressure of `20mm` of `Hg` at temperature T, while pure water exerts a pressure of `20.0126mm Hg` at the same temperature. Calculate the solubility product constant of `A_(3)B_(4)` at the same temperature.

Text Solution

Verified by Experts

The correct Answer is:
`5.4 xx10^(-13)`

`P_(s) = 20 P^(@) = 20.0126`
`(P^(@)-P_(s))/(P^(@)) = (0.0126)/(20) = (n)/(n+N) =(n)/(N)`
`("mole of solute")/("moles of" H_(2)O) = 0.0063`
1 mole `H_(2)O = 18g = 18 mL`
`18 mL` solution `= 0.0063` mole
`1L` solution `= (0.00063)/(18) xx 1000 = 0.35` mole/L
Let solubility of salt `A_(3)B_(4)` is s then
`7s = 0.035`
`s = 0.005` mole/L
`k_(sp) = 3^(3).4^(4) (s)^(7) = 27 xx 256 xx (0.005)^(7)`
`k_(sp) = 5.4 xx 10^(-13)`
Promotional Banner

Similar Questions

Explore conceptually related problems

The lowering of vapour pressure in a saturated aq. Solution of salt AB is found to be 0.108 torr. If vapour pressure of pure solvent at the same temperature is 300 torr, find the solubility product of salt AB.

A solution containing 30g of a nonvolatile solute in exactly 90g water has a vapour pressure of 21.85 mm Hg at 25^(@)C . Further 18g of water is then added to the solution. The resulting solution has vapour pressure of 22.18 mm Hg at 25^(@)C . calculate (a) molar mass of the solute, and (b) vapour pressure of water at 25^(@)C .

The vapour pressure of a certain liquid is given by the equation: Log_(10)P = 3.54595 - (313.7)/(T) +1.40655 log_(10)T where P is the vapour pressure in mm and T = Kelvin Temperature. Determine the molar latent heat of vaporisation as a function of temperature. calculate the its value at 80K .

Two liquids A and B form an ideal solution. At 300 K , the vapour pressure of a solution containing 1 mol of A and 3 mol of B is 550 mm Hg . At the same temperature, if 1 mol more of B is added to this solution, the vapour pressure of the solution increases by 10 mm Hg . Determine the vapour pressure of A and B in their pure states.

Mixture of two liquids A and B is placed in cylinder containing piston. Piston is pulled out isotehrmally so that volume of liquid decreases but that of vapour increases. When negligibly small amount of liquid was remaining the mole fraction of A in vapour is 0.4 . Given P_(A)^(@) = 0.4 atm and P_(B)^(@) = 1.2 atm at the experimental temperature. Calculate the total pressure at whcih the liquid has almost evaporated. (Assume ideal behaviour)

The vapour pressure of pure benzene at a certain temperature is 640 mm Hg . A non-volatile solid weighing 2.175g is added to 39.0g of benzene. The vapour pressure of the solution is 600mm Hg . What is the molar mass of the solid substance?

Two liquids X and Y from an ideal solution at 300K , Vapour pressure of the Solution containing 1 mol of X and 3 mol of Y is 550 mmHg . At the same temperature, if 1 mol of Y is further added to this solution ,vapour pressur of the solutions increases by 100 mmHg Vapour pressure (in mmHg) of X and Y in their pure states will be,respectively

An ideal gas has a specific heat at constant pressure C_P=(5R)/2 . The gas is kept in a closed vessel of volume 0.0083m^3 , at a temperature of 300K and a pressure of 1.6xx10^6 N//m^2 . An amount of 2.49xx10^4 Joules of heat energy is supplied to the gas. Calculate the final temperature and pressure of the gas.