The solubility of AgCl(s) in water at 25 C is 1.33x10^-5 M and its deltaH of solution is 65.7 kJ/mol .
What is its solubility at 60.1 C ?
Please explain how to solve for this!!
To solve for the solubility of AgCl(s) at 60.1°C, we can use the Van't Hoff equation, which relates the change in solubility with temperature to the enthalpy change (deltaH) of the solution. The equation is as follows:
ln(K2/K1) = -(deltaH/R) * ((1/T2) - (1/T1))
Where:
K2 and K1 are the solubility constants (or concentrations) of the substance at temperatures T2 and T1, respectively.
deltaH is the enthalpy change of solution.
R is the gas constant (8.314 J/(mol·K)).
T2 and T1 are the temperatures in Kelvin.
Given information:
T1 = 25°C = 298.15 K
T2 = 60.1°C = 333.25 K
deltaH = 65.7 kJ/mol = 65.7 × 10³ J/mol
We can rearrange the equation to solve for K2:
ln(K2/K1) = -(deltaH/R) * ((1/T2) - (1/T1))
K2/K1 = e^(-(deltaH/R) * ((1/T2) - (1/T1)))
K2 = K1 * e^(-(deltaH/R) * ((1/T2) - (1/T1)))
Now, plug in the given values and calculate:
K1 = 1.33 × 10^(-5) M (Solubility at 25°C)
K2 = (1.33 × 10^(-5) M) * e^(-(65.7 × 10³ J/mol) / (8.314 J/(mol·K)) * ((1/333.25 K) - (1/298.15 K)))
To calculate the solubility at 60.1 °C, simply substitute the value of K2 into the equation.
K2 = 1.33 × 10^(-5) M * e^(-(65.7 × 10³ J/mol) / (8.314 J/(mol·K)) * ((1/333.25 K) - (1/298.15 K)))