Calculate the solubility of silver chloride in a 0.1 mol/L solution of sodium chloride at 25°C.?
At SATP, Ksp AgCl(s) = 1.8 x 10^-10
What did I do wrong?
AgCl <----> Ag+ + Cl-
[Ag+] = x
[Cl-] = x + 0.1
Ksp = [Ag+] [Cl-]
1.8 x 10^-10 = (x) ( x + 0.1)
x = molar solubility = 1.8 x 10^-9 M
I don't see an error. SATP refers to 25o and that's what you have for Ksp. The solubility is certainly smaller than x + 0.1 so x + 0.1 certainly is almost 0.1. The only thing I can think of is that perhaps the problem wanted grams/L instead of molarity. And finally, I don't know the level of the class but sometimes one must make a correction for ionic strength and that can make a SMALL difference.
To calculate the solubility of silver chloride in a 0.1 mol/L solution of sodium chloride at 25°C, you did everything correctly up until the final calculation.
The equation you set up using the solubility product constant (Ksp) for silver chloride is correct:
Ksp = [Ag+] [Cl-]
You correctly identified the concentrations of Ag+ and Cl- as x and x + 0.1, respectively.
However, in the last step, you mistakenly calculated the molar solubility of silver chloride as 1.8 x 10^-9 M. The correct calculation should yield a different result.
To solve for x (the molar solubility), you need to rearrange the equation:
1.8 x 10^-10 = x (x + 0.1)
Now, distribute x into the parenthesis:
1.8 x 10^-10 = x^2 + 0.1x
Rearrange the equation in standard quadratic form:
x^2 + 0.1x - 1.8 x 10^-10 = 0
At this point, you can use the quadratic formula to solve for x. The quadratic formula is:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = 1, b = 0.1, and c = -1.8 x 10^-10. Plug the values into the quadratic formula:
x = (-0.1 ± √(0.1^2 - 4(1)(-1.8 x 10^-10))) / (2(1))
Simplify the equation:
x = (-0.1 ± √(0.01 + 7.2 x 10^-10)) / 2
x = (-0.1 ± √(7.2 x 10^-10)) / 2
Since we are dealing with solubility, x cannot be negative. Take the positive value:
x = (-0.1 + √(7.2 x 10^-10)) / 2
x = (√(7.2 x 10^-10) - 0.1) / 2
Now, calculate the value of x using a calculator. The result will be the solubility of silver chloride in a 0.1 mol/L solution of sodium chloride at 25°C.