How does addition of sodium chloride affect solubility of silver chloride?
Answers
Silver chloride is so insoluble in water (.0.002 g/L) that a saturated solution contains only about 1.3 x 10-5 moles of AgCl per liter of water.
H2O
AgCl(s) <-----> Ag+(aq) + Cl-(aq)
Strict adherence to the rules for writing equilibrium constant expressions for this reaction gives the following result.
equation
(Water isn't included in the equilibrium constant expression because it is neither consumed nor produced in this reaction, even though it is a vital component of the system.)
The [Ag+] and [Cl-] terms represent the concentrations of the Ag+ and Cl- ions in moles per liter when this solution is at equilibrium. The third term--[AgCl]--is more ambiguous. It doesn't represent the concentration of AgCl dissolved in water because we assume that AgCl dissociates into Ag+ ions and Cl- ions when it dissolves in water. It can't represent the amount of solid AgCl in the system because the equilibrium is not affected by the amount of excess solid added to the system. The [AgCl] term has to be translated quite literally as the number of moles of AgCl in a liter of solid AgCl.
The concentration of solid AgCl can be calculated from its density and the molar mass of AgCl.
equation
This quantity is a constant, however. The number of moles per liter in solid AgCl is the same at the start of the reaction as it is when the reaction reaches equilibrium.
Since the [AgCl] term is a constant, which has no effect on the equilibrium, it is built into the equilibrium constant for the reaction.
[Ag+][Cl-] = Kc x [AgCl]
This equation suggests that the product of the equilibrium concentrations of the Ag+ and Cl- ions in this solution is equal to a constant. Since this constant is proportional to the solubility of the salt, it is called the solubility product equilibrium constant for the reaction, or Ksp.
Ksp = [Ag+][Cl-]
The Ksp expression for a salt is the product of the concentrations of the ions, with each concentration raised to a power equal to the coefficient of that ion in the balanced equation for the solubility equilibrium.
Practice Problem 1:
Write the Ksp expression for a saturated solution of CaF2 in water.
Well, let us first consider the solubility equilibrium that governs the dissolution of silver chloride….
AgCl(s)⇌Ag++Cl−
AT ROOM TEMPERATURE. And if you go thru the calculation silver chloride is soluble in the ppm range in PURE WATER…
But the equilibrium expression does not interrogate the source of the chloride ions. And thus if [Cl−] is INCREASED, i.e. by addition of a soluble chloride salt, such as sodium or potassium chloride, or hydrochloric acid, the solubility of the silver salt should be proportionally reduced in either scenario. This is an example of the common ion effect. And thus I think that the premise of the question is mistaken.
It might seem that we could reduce the concentration of silver ion to ANY desired level, simply by ramping up the concentration of the common chloride ion, however, at HIGH concentrations of halide ion, competing equilibria such as the formation of [AgCl2]− might become viable.
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