What mass of silver chromate can be dissolved in 125 mL of water?
How do I find this out?
Ag2CrO4 is not very soluble. Look up the solubility product and calculate the solubility from Ksp.
Wow, thank you so much. I realised that I wrote my chemical formula wrong. I now what to do now. :)
To determine the mass of silver chromate (Ag2CrO4) that can be dissolved in 125 mL of water, you need to use the solubility product constant (Ksp). The Ksp is the equilibrium constant for the dissolution of a solid compound into its constituent ions in a saturated solution.
In this case, you need to look up the solubility product constant (Ksp) for silver chromate. Let's assume that the Ksp value for Ag2CrO4 is 1.8 x 10^-12 (this value may vary depending on the source). The Ksp expression for this compound is:
Ksp = [Ag+]^2[CrO4-]
Since Ag2CrO4 dissociates completely, the concentration of silver ions ([Ag+]) will be twice the concentration of Ag2CrO4 that dissolves:
[Ag+] = 2x
Substituting this into the Ksp expression, we get:
1.8 x 10^-12 = (2x)^2[CrO4-]
Now, let's assume that x is the concentration of Ag2CrO4 that dissolves. Therefore, the concentration of silver ions ([Ag+]) is 2x.
Next, you need to convert the volume of water from mL to L, since Ksp is typically expressed in terms of Molarity (mol/L).
125 mL of water is equal to 0.125 L.
Now you can set up the equation to solve for x:
1.8 x 10^-12 = (2x)^2 * x
Simplifying the equation further gives:
1.8 x 10^-12 = 4x^3
To solve for x, take the cube root of both sides:
x = (1.8 x 10^-12 / 4)^(1/3)
Now, plug the values into a calculator to find the value of x. In this case, x represents the concentration of Ag2CrO4 that dissolves.
Once you know the concentration of Ag2CrO4, you can calculate the mass of silver chromate that dissolves in 125 mL of water using the formula:
mass = concentration x volume.
Therefore, the mass of silver chromate that can be dissolved in 125 mL of water can now be determined using the calculated concentration of Ag2CrO4 and the volume of water.