Calculate solubility of AgCl (in g/L) in a 6.5*10^-3 M silver Nitrate solution ksp= 1.8 * 10^-10
(AgCl molecular mass is 143.3g) any idea on how to do this this is what i tried>
AgCl--> Ag + Cl
I 6.5*10^-3 0
C same x
E same x
ksp = [Ag][Cl]
(1.8 * 10^-10) = (6.5 * 10^-3)x
x = 2.77 * 10^-8 M
I don't know what to do next please help me thank you
see your other post above.
To calculate the solubility of AgCl in a 6.5*10^-3 M silver nitrate solution, you can use the concept of the solubility product constant (Ksp). The Ksp is the equilibrium constant for the dissociation of a sparingly soluble salt, in this case, AgCl.
The chemical equation for the dissociation of AgCl is:
AgCl ↔ Ag+ + Cl-
In the given 6.5*10^-3 M silver nitrate solution, we assume that both the Ag+ and Cl- ions come from the dissociation of AgNO3. Therefore, the initial concentration of Ag+ ions is also 6.5*10^-3 M.
Now, you correctly set up an ICE table to determine the equilibrium concentration of Cl-:
AgCl ↔ Ag+ + Cl-
I 6.5*10^-3 0
C same x
E same x
Since the stoichiometry of the balanced chemical equation is 1:1, the concentration of Cl- ions is also x.
Now, you can use the solubility product constant equation to calculate the value of x:
Ksp = [Ag+][Cl-]
Substituting the known values:
(1.8*10^-10) = (6.5*10^-3)(x)
Rearranging the equation to solve for x:
x = (1.8*10^-10)/(6.5*10^-3)
Evaluating the expression:
x ≈ 2.77*10^-8 M
So the solubility of AgCl in a 6.5*10^-3 M silver nitrate solution is approximately 2.77*10^-8 M.
To convert this solubility into grams per liter (g/L), you can use the molar mass of AgCl.
The molar mass of AgCl is given as 143.3 g/mol.
Convert the solubility to moles per liter:
2.77*10^-8 mol/L
Now, multiply the moles per liter by the molar mass:
(2.77*10^-8 mol/L) * (143.3 g/mol)
Calculating the result:
≈ 3.97*10^-6 g/L
Therefore, the solubility of AgCl in a 6.5*10^-3 M silver nitrate solution is approximately 3.97*10^-6 g/L.