A 200.0mL solution with 3.1 x 10^-3
mol of MA will begin to precipitate at 19.4 C. Calculate each of
the following at this temperature.
a) Ksp
b)change in G
A. I assume M^+ and A^- but it could be M^2+ and A^2- etc.
(MA) = 0.0031 x (1000 mL/200 mL) = 0.0031*5 = ?
Ksp = (M^+)(A^-) = (0.0031*5)^2
B. dG = -RT*lnK
To calculate the Ksp (solubility product constant) and change in Gibbs free energy, we need additional information such as the balanced chemical equation for the precipitation reaction and the molar solubility of the precipitate (MA).
Please provide the balanced equation and the molar solubility (if available) so that I can assist you further.
To answer these questions, we need to know the equilibrium equation for the precipitation reaction and the appropriate values for the reaction.
a) Ksp, also known as the solubility product constant, represents the equilibrium constant for the dissolution of a sparingly soluble salt in water. The equilibrium equation for this precipitation reaction can be written as:
MA (s) ⟷ M+ (aq) + A- (aq)
The solubility product constant expression for this reaction is:
Ksp = [M+]^x [A-]^y
Where [M+] represents the concentration of the cation (M+) in the solution and [A-] represents the concentration of the anion (A-) in the solution. The values of x and y represent the stoichiometric coefficients of the ions in the balanced equation.
Since the problem statement indicates that MA is a sparingly soluble salt, it means that the concentration of the ions resulting from its dissolution will be very low and can be assumed to be the same as the initial concentration of MA.
Given that the initial concentration of MA is 3.1 x 10^-3 mol in a 200.0 mL solution, we can calculate the concentration of M+ and A- by dividing this value by the volume of the solution:
MA concentration = (3.1 x 10^-3 mol) / (200.0 mL) = 1.55 x 10^-5 mol/mL
Therefore, [M+] = [A-] = 1.55 x 10^-5 mol/mL. The stoichiometric coefficient for both ions in this equation is 1, so we can rewrite the solubility product constant expression as:
Ksp = (1.55 x 10^-5)^2 = 2.4025 x 10^-10 (mol/mL)^2
b) The change in Gibbs free energy (ΔG) can be calculated using the equation:
ΔG = -RTlnK
Where R is the gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin, and lnK is the natural logarithm of the equilibrium constant (K).
To convert the temperature from Celsius to Kelvin, we can use the equation:
T(K) = T(°C) + 273.15
Given that the temperature is 19.4 °C, the temperature in Kelvin becomes:
T(K) = 19.4 + 273.15 = 292.55 K
Now, we can substitute the values into the equation to calculate the change in Gibbs free energy:
ΔG = - (8.314 J/(mol·K)) * 292.55 K * ln(2.4025 x 10^-10)
Calculating this expression using a scientific calculator, we can determine the value of ΔG.
Please note that we used the natural logarithm (ln) in this equation instead of the common logarithm (log) because the equilibrium constant (K) is defined using the natural logarithm.