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tatiyna
3 years ago
11

A generic salt, AB, has a molar mass of 241 g/mol and a solubility of 2.60 g/L at 25 °C. What is the Ksp of this salt at 25 °C?

Chemistry
1 answer:
alekssr [168]3 years ago
3 0
When the salt dissolves, it dissociates as follows;
AB --> A⁺ + B⁻
The molar solubility is amount of moles of the salt that can be dissolved in 1 L of solution.
solubility of the salt is - 2.60 g/L, to find molar solubility we have to multiply by the molar mass.
2.60 g/L /241 g/mol = 0.0108 mol/L
ksp = [A⁺][B⁻]
molar solubility of salt is equal to the molar solubility of cation and anion 
ksp = (0.0108)(0.0108)
ksp = 1.17 x 10⁻⁴
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The question is incomplete, here is the complete question:

The equilibrium constant for the reaction

N₂O₄(g)⇌2NO₂ at 2°C is Kc = 2.0

If each yellow sphere represents 1 mol of N₂O₄(g) and each gray sphere 1 mol of NO₂ which of the following 1.0 L containers represents the equilibrium mixture at 2°C?

The image is attached below.

<u>Answer:</u> The system which represents the equilibrium having value of K_c=2.0 is system (b)

<u>Explanation:</u>

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K_c

For a general chemical reaction:

aA+bB\rightarrow cC+dD

The expression for K_{c} is written as:

K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}

For the given chemical equation:

N_2O_4(g)\rightleftharpoons 2NO_2

The expression of K_c for above equation follows:

K_c=\frac{[NO_2]^2}{[N_2O_4]}      .......(1)

We are given:

Volume of the container = 1.0 L

Value of K_c = 2.0

Molarity of the substance is calculated by using the equation:

\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume}}

For the given images:

  • <u>For a:</u>

Number of Gray spheres = 8 moles

Number of yellow spheres = 4 moles

Putting values in expression 1, we get:

K_c=\frac{(8/1)^2}{(4/1)}\\\\K_c=16

  • <u>For b:</u>

Number of Gray spheres = 4 moles

Number of yellow spheres = 8 moles

Putting values in expression 1, we get:

K_c=\frac{(4/1)^2}{(8/1)}\\\\K_c=2

  • <u>For c:</u>

Number of Gray spheres = 6 moles

Number of yellow spheres = 6 moles

Putting values in expression 1, we get:

K_c=\frac{(6/1)^2}{(6/1)}\\\\K_c=6

  • <u>For d:</u>

Number of Gray spheres = 2 moles

Number of yellow spheres = 8 moles

Putting values in expression 1, we get:

K_c=\frac{(2/1)^2}{(8/1)}\\\\K_c=\frac{1}{2}

Hence, the system which represents the equilibrium having value of K_c=2.0 is system (b)

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