An AgNO3 solution (44 mL/0.44 M) has been mixed with an NaCl solution (83 mL/1.35x 10^-2 M)
a)Determine the ion product of the precipitate(potential precipitate)
b) Would a precipitate be created? The Ksp of AgCl(s) proves to be 1.8 x 10^-10
If you will follow the instructions you will work the problem.
a). Do the ion product
b). compare the ion product with Ksp. Post your work if you get stuck.
How much heat (in Joules) is gained by the water where a chemical reaction takes place in 100 mL aqueous solution and has a temperature increase of 12C?
To determine the ion product of the potential precipitate, we first need to write the balanced chemical equation for the reaction between AgNO3 and NaCl:
AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
From the equation, we can see that AgCl is the potential precipitate.
a) The ion product, also known as the solubility product constant (Ksp), represents the concentration of the ions in the saturated solution at equilibrium. In this case, the Ksp for AgCl is given as 1.8 x 10^-10.
The dissociation of AgCl can be represented as follows:
AgCl(s) ⇌ Ag+(aq) + Cl-(aq)
Using the stoichiometry of the balanced equation, we can determine the concentration of Ag+ and Cl- ions in the solution using the molarity and volume data given for AgNO3 and NaCl solutions, respectively.
For AgNO3:
Volume (V) = 44 mL = 0.044 L
Molarity (M) = 0.44 M
moles of Ag+ = Molarity x Volume = 0.44 M x 0.044 L = 0.01936 mol
For NaCl:
Volume (V) = 83 mL = 0.083 L
Molarity (M) = 1.35 x 10^-2 M
moles of Cl- = Molarity x Volume = 1.35 x 10^-2 M x 0.083 L = 0.0011155 mol
Therefore, the concentration of Ag+ ions is 0.01936 mol/L, and the concentration of Cl- ions is 0.0011155 mol/L.
Now, to calculate the ion product (Ksp), we multiply the concentrations of Ag+ and Cl- ions:
Ksp = [Ag+][Cl-] = (0.01936 mol/L) x (0.0011155 mol/L) = 2.1594 x 10^-5
Hence, the ion product for the potential precipitate, AgCl, is 2.1594 x 10^-5.
b) In order for a precipitate to form, the ion product (Qsp) must exceed the solubility product constant (Ksp). If Qsp > Ksp, then a precipitate will form.
In this case, Qsp is 2.1594 x 10^-5, and Ksp is 1.8 x 10^-10. Since Qsp > Ksp, a precipitate of AgCl will be created.