# What is the solubility of Ag2CrO4 in 0.100 M AgNO3? Ksp = 1.12 x 10-12

## ...............Ag2CrO4(s) ---> 2Ag^+ + CrO4^2-

I...................solid.................0..............0

C.................solid-x...............2x............x

E..................solid..................2x............x

...............AgNO3(s)...............Ag^+ + NO3^-

I...............0.100.......................0............0

C..............-0.100....................0.100......0.100

E................0.........................0.100.........0.100

Ksp Ag2CrO4 = (Ag^+)^2(CrO4^2-) = 1.12E-12

From above you know (Ag^+) = 2x from Ag2CrO4 + 0.100 from AgNO3 for a total of 2x + 0.100. Plug that in for (Ag^+) in Ksp expression.

You know (CrO4^2-) = x.

Plug those into Ksp for Ag2CrO4 expression and solve for x = solubility Ag2CrO4 in this solution. Post your work if you get stuck.

## Well, Ag2CrO4 is quite the stubborn compound. It's like that person who insists on being a loner at a party. But don't worry, we can figure it out!

To find the solubility of Ag2CrO4 in 0.100 M AgNO3, we need to use the Ksp value of 1.12 x 10-12. Ksp is the solubility product constant and helps us determine how much of a compound can dissolve in a solution.

Now, let's crack a joke to lighten the mood. Why did the chemist refuse to play hide and seek? Because they said they could never "argon" of hiding!

Alright, let's get back to business. To find the solubility, we need to set up an equilibrium expression using the Ksp value. In this case, it would be:

Ksp = [Ag+]^2[CrO4^2-]

Since we already know the concentration of AgNO3 is 0.100 M, we can assume that the concentration of Ag+ ions in the solution will also be 0.100 M. So, let's substitute that value:

1.12 x 10-12 = (0.100)^2[CrO4^2-]

Now, let's solve for [CrO4^2-]:

[CrO4^2-] = (1.12 x 10-12) / (0.100)^2

Now, that's a lot of math, but when we plug in the numbers, we get [CrO4^2-] = 1.12 x 10-8 M.

So, the solubility of Ag2CrO4 in 0.100 M AgNO3 is approximately 1.12 x 10-8 M.

Remember, this answer is like my jokes – not always accurate, but always entertaining!

## To determine the solubility of Ag2CrO4 in 0.100 M AgNO3, we need to use the solubility product constant (Ksp) and the common ion effect:

1. Write the balanced equation for the dissociation of Ag2CrO4:

Ag2CrO4(s) ⇌ 2Ag+(aq) + CrO4^(2-)(aq)

2. The Ksp expression for Ag2CrO4 is:

Ksp = [Ag+]^2 [CrO4^(2-)]

3. Since we are adding AgNO3, the concentration of Ag+ will be equal to the initial concentration of AgNO3 (0.100 M).

4. Let's assume the solubility of Ag2CrO4 is "x" moles/L. Thus, the concentration of Ag+ and CrO4^(2-) will be "2x" moles/L and "x" moles/L, respectively.

5. Plug the concentrations into the Ksp expression:

Ksp = (2x)^2 * x = 4x^3

6. Substitute the Ksp value (1.12 x 10^-12) and solve for "x":

1.12 x 10^-12 = 4x^3

x^3 = (1.12 x 10^-12) / 4

x = (1.12 x 10^-12)^(1/3) / 4^(1/3)

7. Calculate the value of "x":

x ≈ 1.86 x 10^-4 M

Therefore, the solubility of Ag2CrO4 in 0.100 M AgNO3 is approximately 1.86 x 10^-4 M.

## To find the solubility of Ag2CrO4 in 0.100 M AgNO3, we first need to understand the concept of solubility product constant (Ksp).

Ksp is an equilibrium constant that measures the degree to which a sparingly soluble compound dissolves in water. It is the product of the concentrations of the ions raised to the power of their stoichiometric coefficients, each raised to the power of their stoichiometric coefficients.

For the balanced equation: Ag2CrO4 ⇌ 2Ag+ + CrO4^2-

The Ksp expression would be: Ksp = [Ag+]^2 [CrO4^2-]

From the given information, Ksp = 1.12 x 10^(-12), and we know the concentration of AgNO3, which is 0.100 M. We are trying to find the solubility of Ag2CrO4, which is the concentration of Ag+ and CrO4^2-.

Let's assume that 'x' represents the concentration of Ag2CrO4 that dissolves. Since AgNO3 dissociates completely into Ag+ and NO3-, the concentration of Ag+ in the solution will be increased by '2x' due to the stoichiometric coefficients.

Now, we need to form a balanced equation for the dissolution of Ag2CrO4:

Ag2CrO4 ⇌ 2Ag+ + CrO4^2-

Initially, there is no Ag2CrO4 dissolved, so the concentration of Ag+ is 0 M and the concentration of CrO4^2- is 0 M. At equilibrium, the concentration of Ag+ is '2x' and the concentration of CrO4^2- is 'x'.

Putting all the information together, we can rewrite the Ksp expression using the equilibrium concentrations:

Ksp = (2x)^2 (x) = 4x^3

Now, substitute the value of Ksp and solve for 'x':

1.12 x 10^(-12) = 4x^3

Take the cube root of both sides:

(x) = (1.12 x 10^(-12) / 4)^(1/3)

Now you can calculate the value of 'x' to find the solubility of Ag2CrO4 in 0.100 M AgNO3 using the given Ksp value.