10.0 mL of 0.05 M AgNO3 and 10.0 mL 0.05 M NaCl are combined in a test tube and a precipitate of AgCl forms. After the precipitate is centrifuged down, the remaining solution (called the supernatant) is tested for the concentration of Ag+ ion. The concentration is found to be 1.3x10^-5. Calculate the Ksp of AgCl from this information. Show work please. Thanks!
...AgCl(s) ==> Ag^+(aq) + Cl^-(aq)
I..solid.......0...........0
E..solid.....1.35E-5.....1.35E-5
Ksp = (Ag^+)(Cl^-) = (1.35E-5)^2 = ?
To calculate the Ksp (solubility product constant) of AgCl from the given information, we will use the following equation:
AgCl(s) ⇌ Ag+(aq) + Cl-(aq)
We know that the concentration of Ag+ ions in the supernatant is 1.3x10^-5 M. Since AgNO3 and NaCl are both strong electrolytes and completely ionize in water, we can assume that the concentration of Cl- ions in the supernatant is also 1.3x10^-5 M.
Let's assume that the concentration of Ag+ ions is x M. Since AgCl dissolves in water according to the balanced equation, the concentration of Cl- ions will also be x M.
Using the Ksp expression for AgCl, we have:
Ksp = [Ag+][Cl-]
Substituting the known concentrations, we get:
Ksp = (x)(x)
= x^2
Since the concentration of Ag+ ions and Cl- ions are both x, we can substitute 1.3x10^-5 M for x:
Ksp = (1.3x10^-5)^2
Calculating this, we get:
Ksp = 1.69x10^-10
Therefore, the solubility product constant (Ksp) of AgCl is equal to 1.69x10^-10.
To find the value of the solubility product constant (Ksp) for AgCl, you can use the given concentration of Ag+ ion in the supernatant.
First, let's write the balanced chemical equation for the precipitation of AgCl:
AgNO3 + NaCl -> AgCl + NaNO3
From the equation, we know that 1 mole of AgNO3 reacts with 1 mole of NaCl to form 1 mole of AgCl.
Given:
Initial volume of AgNO3 = 10.0 mL = 0.010 L
Initial concentration of AgNO3 = 0.05 M
Initial moles of AgNO3 = Initial concentration x Initial volume
= 0.050 mol/L x 0.010 L
= 0.0005 mol
Similarly, the initial moles of NaCl = 0.0005 mol.
From the balanced equation, we know that the moles of AgCl formed will be equal to the moles of AgNO3 and NaCl. Therefore, the moles of AgCl formed = 0.0005 mol.
The volume of the supernatant is the sum of the volumes of AgNO3 and NaCl, which is 10.0 mL + 10.0 mL = 20.0 mL = 0.020 L.
The concentration of Ag+ ion in the supernatant is given as 1.3x10^-5 M.
The concentration of Ag+ can be calculated as the moles of Ag+ ion divided by the volume of the supernatant:
Concentration of Ag+ = Moles of Ag+ / Volume of supernatant
= 0.0005 mol / 0.020 L
= 0.025 M
Now, we can set up the expression for Ksp using the concentration of Ag+ ion:
Ksp = [Ag+][Cl-]
Since the salt AgCl has a 1:1 stoichiometry, the concentration of Cl- ions is equal to the concentration of Ag+ ions.
Therefore, Ksp = (concentration of Ag+) x (concentration of Cl-)
= (0.025 M) x (0.025 M)
= 6.25 x 10^-4
Therefore, the solubility product constant (Ksp) for AgCl is 6.25 x 10^-4.