Question

The osmotic pressure of a saturated solution of strontium sulfate at 25 ∘C∘C is 21 torrtorr. Part A What is the solubility product of this salt at 25 ∘C∘C ?

Answer 1

Answer

solubility product = 3.18x 10^-7

Explanation:

We were given the pressure in torr then we need to convert to atm for consistency, ten we have

21torr/760= 0.0276315789 atm

21 Torr = .0276315789 atm

P = i M S T

M = P / iRT

Where p is osmotic pressure

T= temperature= 25C+ 273= 298K

for XY vanthoff factor i = 2

S = 0.0821 L-atm / mol K

M = .0276315789 atm / (2)(0.0821 L atm / K mole)(298 K)

M = 0.000564698046 mol/liters

solubility= 0.000564698046 mol/liters

Ksp = [X+][Y-]

Ksp = X^2

Ksp = [Sr^+2] * [SO4^-2]

Ksp = X^2

Ksp = (0.000564698046)^2

Ksp = 3.18883883 × 10-7

Ksp = 3.18x 10^-7

solubility product = 3.18x 10^-7

Therefore, the solubility product of this salt at 25 ∘C∘C is 3.18x 10^-7

Answer 2

Osmotic pressure has been the pressure exerted to the liquid in a system to avoid the backflow. The solubility product of the given salt is $1.27\;\times\;10^{-6}$.

What is a solubility product?

The solubility product has been given as the equilibrium constant for the salt at the specified temperature.

The solubility product (ksp) for a salt can be given as:

$ksp=\rm [Sr^{2+}][SO_4^{2-}]$

The concentration (M) can be given as:

$\pi=MRT$

Where the osmotic pressure is

$\pi = 21\; \rm torr\\\\\pi=\dfrac{21}{760}\;atm$

The temperature is,

$T=25^\circ \text C\\T=298\;\text K$

The gas constant has the value of,

$R=0.0821\;\rm atm.L/mol.K$

Substituting the values for the concentration:

$\dfrac{21}{760}=M\;\times\;0.0821\;\times\;298\\\\M= 0.00111\;M$

The concentration of the ions of salt is 0.0011 M. The solubility product can be given as:

$ksp=[0.0011]^2\\ksp=1.27\;\times\;10^{-6}$

The solubility product of the strontium salt is $1.27\;\times\;10^{-6}$.

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