What is the solubility of silver chromate in pure water
...........Ag2CrO4 ==> 2Ag^+ + CrO4^2-
I..........solid......0.......0
C..........solid......2x.......x
E..........solid......2x.......x
Substitute the E line into the Ksp expression and solve for x
The solubility of silver chromate in pure water can be calculated using the solubility product expression (Ksp) for the compound. The Ksp is the equilibrium constant for the dissociation of the compound into its constituent ions in a saturated solution.
The balanced equation for the dissociation of silver chromate (Ag2CrO4) can be written as follows:
Ag2CrO4 ⇌ 2Ag+ + CrO4^2-
The solubility product expression for this reaction is:
Ksp = [Ag+]^2 * [CrO4^2-]
To determine the solubility of silver chromate, we need to know the value of the Ksp. The calculated value of Ksp for silver chromate at 25°C is approximately 9.0 x 10^-12.
By knowing the Ksp value, we can calculate the solubility of the compound by assuming that the dissociation is complete and using the concentrations of the ions in the equilibrium expression.
Since the stoichiometry of the dissociation reaction is 1:2, the concentration of silver ions is twice that of chromate ions in the saturated solution.
Let's denote the solubility of silver chromate as "s". Therefore, the concentration of silver ions ([Ag+]) is 2s, and the concentration of chromate ions ([CrO4^2-]) is s.
Plugging these values into the Ksp expression, we have:
9.0 x 10^-12 = (2s)^2 * (s)
Simplifying the equation, we get:
4s^3 = 9.0 x 10^-12
Solving for "s", we find:
s ≈ 1.3 x 10^-4
Therefore, the solubility of silver chromate in pure water is approximately 1.3 x 10^-4 moles per liter.
To determine the solubility of silver chromate in pure water, you can refer to solubility product constant (Ksp) values. The Ksp value gives you an idea of how soluble a compound is in water, as it represents the product of the concentrations of the dissolved ions raised to the power of their stoichiometric coefficients.
For silver chromate (Ag2CrO4), the solubility equilibrium is given by the chemical equation:
Ag2CrO4 (s) ⇌ 2Ag+ (aq) + CrO42- (aq)
The solubility product constant expression for this equilibrium is:
Ksp = [Ag+]² [CrO4²⁻]
The solubility product constant for silver chromate is reported to have a value of approximately 1.1 x 10⁻¹² at 25°C. This means that at equilibrium, the silver ion concentration squared multiplied by the chromate ion concentration should equal 1.1 x 10⁻¹².
Therefore, from this information, we can conclude that the solubility of silver chromate in pure water is quite low.