The solubility of AgCl(s) in water at 25 C is 1.33x10^-5 M and its delta H of solution is 65.7 kJ/mol .
What is its solubility at 60.1 C ?
To find the solubility of AgCl(s) in water at 60.1 °C, we can use the van 't Hoff equation:
ln(K2/K1) = (-ΔH/R) * (1/T2 - 1/T1)
Where:
- ΔH is the enthalpy of solution (65.7 kJ/mol)
- R is the gas constant (8.314 J/(mol·K))
- T1 is the initial temperature (25 °C + 273.15 = 298.15 K)
- T2 is the final temperature (60.1 °C + 273.15 = 333.25 K)
- K1 is the solubility at the initial temperature (given as 1.33x10^-5 M)
- K2 is the solubility at the final temperature (what we're trying to find)
First, we need to convert the initial solubility value from Molarity (M) to moles per liter (mol/L):
K1_initial = 1.33x10^-5 M
Next, we can substitute all the known values into the equation:
ln(K2/K1) = (-65.7 kJ/mol / (8.314 J/(mol·K))) * (1/333.25 K - 1/298.15 K)
Now, we can solve for ln(K2/K1):
ln(K2/K1) = (-7.92) * (0.0031803 - 0.0033587)
= (-7.92) * (-0.0001784)
= 0.0014121
To find K2, we can take the exponent of both sides:
K2/K1 = e^(0.0014121)
Now, we can solve for K2:
K2 = K1 * e^(0.0014121)
Substituting the initial solubility value into the equation:
K2 = 1.33x10^-5 M * e^(0.0014121)
Calculating this value will give us the solubility of AgCl(s) in water at 60.1 °C.