calculate the solubility of SrF2 (ksp=7.9E-10) in 0.100 M Sr(NO3)2 ?

The answer is 4.44E-5 M how do you get that?

I Have no clue?

what is the final answer?

i still need help.

Follow up on your later post but the answer really is 4.44E-5M. You just didn't set it up right. Probably didn't square F^- OR didn't include the 2 which squared is 4.

To calculate the solubility of SrF2 in 0.100 M Sr(NO3)2, we need to make use of the concept of common ion effect and the equilibrium expression known as the solubility product constant (Ksp).

1. Write the chemical equation:
SrF2(s) ⇌ Sr2+(aq) + 2F-(aq)

2. Write the Ksp expression:
Ksp = [Sr2+][F-]^2 = 7.9E-10

3. Introduce the common ion effect:
Since we have Sr(NO3)2, which ionizes to produce additional Sr2+ ions, we need to consider the concentration of Sr2+ already present in the solution as a result of the soluble salt.

Let's say the solubility of SrF2 is 'x', then the concentration of Sr2+ due to the dissolved SrF2 is also 'x'. In addition, we have 0.100 M Sr(NO3)2, which adds an extra concentration of Sr2+.

So, the total concentration of Sr2+ ions is 0.100 M + x, which we will use in the Ksp expression.

4. Plug in the values and solve for 'x':
Ksp = (0.100 + x)(x)^2 = 7.9E-10

Solving this quadratic equation can be a bit complex. However, in this case, the Ksp value is very small (10^-10), indicating that the value of 'x' will be much smaller than 0.100 M. We can assume that the addition of 'x' (the solubility of SrF2) to the initial 0.100 M concentration of Sr(NO3)2 will have a negligible impact on the concentration of Sr2+ ions from Sr(NO3)2.

Therefore, we can simplify the equation to:
Ksp = (0.100)(x)^2 = 7.9E-10

Taking the square root of both sides gives:
(0.100)(x) = √(7.9E-10)
x ≈ √(7.9E-10)/0.100

Calculating the approximate value gives:
x ≈ 2.8064E-5

So, the solubility of SrF2 in 0.100 M Sr(NO3)2 is approximately 2.8064E-5 M or 4.44E-5 M (rounded to three significant figures).

This is a Ksp problem with a common ion add on.

Two equilibria. Sr(NO3)2 is a strong electrolyte and ionizes 100% in solution.
..........Sr(NO3)2 ==> Sr^2+ + 2NO3^-
I..........0.100........0........0
C.........-0.100.......0.100...0.200
E...........0..........0.100...0.200

SrF2 is a slightly soluble electrolyte and only dissolves partially. Ksp is in charge.
.........SrF2 --> Sr^2+ + 2F^-
I........solid....0.......0
C........solid....x.......2x
E........solid....x.......2x

so Ksp expression is
Ksp = (Sr^2+)(F^-)^2
Substitute as follows:
(Sr^2+) = x from the SrF2 and 0.1 for the Sr(NO3)2 salt = x+0.1
(F^-) = 2x
and solve for x.
Note: It isn't necessary to solve a full quadratic equation since x + 0.1 for all practical purposes is 0.1.