Calculate the solubility of CaCrO4 if its Ksp is 1.6 x 10-9 at 25°C, in mg/L.Ksp = 1.6 x 10-9
.........CaCrO4 ==> Ca^2+ + CrO4^2-
I....................0........0
C....................x........x
E....................x........x
Substitute into the Ksp expression and solve for x then convert to mg/L.
@DrBob222
How do u convert to mg/l?
To calculate the solubility of CaCrO4, we first need to convert the Ksp value from scientific notation to decimal form.
Ksp = 1.6 x 10^(-9)
Now, let's assume that the solubility of CaCrO4 is "s" moles per liter. Therefore, the dissolution reaction can be represented as:
CaCrO4(s) ⇌ Ca^2+(aq) + CrO4^2-(aq)
The solubility product expression for this reaction is:
Ksp = [Ca^2+][CrO4^2-]
Since the stoichiometric coefficients for CaCrO4, Ca^2+, and CrO4^2- are all 1, the concentration of Ca^2+ and CrO4^2- in terms of "s" can be considered equal. Therefore:
[Ca^2+] = [CrO4^2-] = s
Substituting these values into the solubility product expression:
Ksp = s * s
1.6 x 10^(-9) = s^2
To solve for "s", we can take the square root of both sides of the equation:
s = √(1.6 x 10^(-9))
s = 1.26 x 10^(-4) M
Finally, to convert the solubility from moles per liter (M) to milligrams per liter (mg/L), we need to consider the molar mass of CaCrO4.
The molar mass of CaCrO4 (calcium chromate) is 156.15 g/mol.
Converting the solubility to milligrams per liter:
s (mg/L) = 1.26 x 10^(-4) M * 156.15 g/mol * 1000 mg/g
s (mg/L) = 19.6 mg/L
Therefore, the solubility of CaCrO4 is approximately 19.6 mg/L.
To calculate the solubility of CaCrO4, we need to use the Ksp (solubility product constant) and its equation. The Ksp value is given as 1.6 x 10^(-9).
The general equation for the solubility product constant is as follows:
CaCrO4 (s) ⇌ Ca^2+ (aq) + CrO4^2- (aq)
In this equation, CaCrO4 is the solid compound, and Ca^2+ and CrO4^2- are its constituent ions when it dissolves.
The solubility product expression for this equation can be written as follows:
Ksp = [Ca^2+][CrO4^2-]
To find the solubility (in terms of molar concentration), we usually assume that the solubility of the solid compound "CaCrO4" is "s", which means that the molar concentration of Ca^2+ and CrO4^2- is also "s".
Therefore, we can rewrite the equation as:
Ksp = s^2
Given that Ksp = 1.6 x 10^(-9), we can solve for "s" by taking the square root of Ksp:
s = √(Ksp)
s = √(1.6 x 10^(-9))
s ≈ 1.26 x 10^(-5) M
To convert the solubility from moles per liter (M) to milligrams per liter (mg/L), you need to know the molar mass of CaCrO4. Calcium chromate (CaCrO4) has a molar mass of 152.1 g/mol.
To convert from molar concentration (M) to milligrams per liter (mg/L), you can use the following conversion factor:
1 M = 1000 mg/L
Thus, to calculate the solubility in mg/L, we multiply the molar concentration by the molar mass and then convert to mg/L:
Solubility = (1.26 x 10^(-5) M) * (152.1 g/mol) * (1000 mg/g)
Solubility ≈ 1.92 mg/L
Therefore, the solubility of CaCrO4 is approximately 1.92 mg/L when its Ksp is 1.6 x 10^(-9) at 25°C.