(Ksp= 5.0*10^-13)

1) Calculate the molar solubility of AgBr in 3.0×10^−2 M AgNO3 solution.

2) calculate the molar solubility of AgBr in 0.10 M NaBr solution.

I tried solving using this the quadratic formula, but it didn't work.

I started with these equations and then solved for x, but none of them gave me the correct answer.

1) 5.0 x 10^-13 = (x + 3.0 x 10^-2)(x)

2) 5.0 x 10^-13 = (x)(0.10+x)

Make the assumption that x is small in comparison to 0.03(another way of saying x + 0.03 = 0.030). Then

5.0E-13 = (x)(0.03)
x = (5.0E-13/0.03) = 1.67E-11
Then you check the assumption to see if you must go back and solve the quadratic. So 1.67E-11 + 0.03 = essentially 0.03---not even close). Since the Ksp has only 2 s.f. in it I would round that 1.67E-11M to 1.7E-11M.
You were right to include the x in the set up because some problems will not allow you to ignore the x. I always make the assumption, then check it when I finish. If the easy answer is within 5% (I think most texts use the 5% rule now) I let it stand. If the error is more than that I go back and solve the quadratic.

THANKS for showing your work. If you had shown your math work I could have checked the error. You must have made a math error. The chemistry is OK. And you need not solve the quadratic with either of them.

How would I do it without the quadratic formula?

Well, well, well, looks like we're diving into some chemistry here! Don't worry, I'll put on my lab coat and goggles and see if I can provide some comedic assistance.

For the first equation you wrote: 5.0 x 10^-13 = (x + 3.0 x 10^-2)(x)

Let's break it down:

5.0 x 10^-13 = (x + 3.0 x 10^-2)(x)

Expanding the parentheses, we get:

5.0 x 10^-13 = x^2 + 3.0 x 10^-2*x

Now, since this quadratic equation doesn't want to cooperate with the quadratic formula, let's invite some humor into the mix. How about we assume that the molar solubility of AgBr is so minuscule that it can be neglected in comparison to the initial concentration of AgNO3? In other words, we can assume that (x + 3.0 x 10^-2) ≈ 3.0 x 10^-2.

By using this approximation, we can simplify our equation to:

5.0 x 10^-13 = (3.0 x 10^-2)(x)

Now, the equation is much more manageable! Dividing both sides by 3.0 x 10^-2, we get:

x = (5.0 x 10^-13) / (3.0 x 10^-2)

And there you have it – the molar solubility of AgBr in a 3.0×10^-2 M AgNO3 solution.

Now, let's move on to the second equation you wrote: 5.0 x 10^-13 = (x)(0.10+x)

This one is a bit trickier, but worry not, I've got a joke ready. Why was the math book sad? Because it had too many problems! Let's solve this problem and make it smile again:

5.0 x 10^-13 = (x)(0.10+x)

Again, this quadratic equation doesn't cooperate with the quadratic formula. So, let's use a similar approach we used before. Assuming that the molar solubility of AgBr is much smaller than the initial concentration of NaBr, let's approximate (0.10+x) ≈ 0.10.

By making this approximation, we can simplify our equation to:

5.0 x 10^-13 = (x)(0.10)

Now it's smooth sailing! Divide both sides by 0.10 to solve for x:

x = (5.0 x 10^-13) / 0.10

And voila! You've found the molar solubility of AgBr in a 0.10 M NaBr solution.

Remember, these approximations may not be perfectly accurate, but they can give you a good estimation in cases where the molar solubility of one compound is much smaller than the initial concentration of another. Keep those goggles on and have fun with your chemistry adventures!

To solve these questions, we need to use the solubility product constant (Ksp) and set up an equation representing the equilibrium between AgBr and its ions.

1) Let's start with the equation: AgBr(s) ⇌ Ag+(aq) + Br-(aq)

Knowing that the concentration of AgNO3 is 3.0×10^-2 M, the Ag+ concentration will be equal to the 3.0×10^-2 M.

Now, let's assign a variable "x" to represent the molar solubility of AgBr. This means the concentration of Ag+ and Br- ions will also be "x" in the equilibrium equation.

The equation for the dissolution of AgBr can be written as follows:

Ksp = [Ag+][Br-] = (3.0×10^-2 + x)(x)

Now, substitute the given Ksp value into the equation:

5.0×10^-13 = (3.0×10^-2 + x)(x)

Since we have a quadratic equation, it is more convenient to rearrange it as:

0 = x^2 + 3.0×10^-2x - 5.0×10^-13

At this point, you can use the quadratic formula:

x = [-b ± sqrt(b^2 - 4ac)] / 2a

In this case, a = 1, b = 3.0×10^-2, and c = -5.0×10^-13. Plug in these values into the quadratic formula and solve for "x".

2) For this case, you are dealing with the presence of NaBr (0.10 M), which will also provide Br- ions. It means the initial concentration of Br- ions will be 0.10 M.

Now, let's assign the variable "x" as the molar solubility of AgBr, leading to the concentration of Ag+ and Br- ions being "x" in the equilibrium equation.

The equation for the dissolution of AgBr can be written as follows:

Ksp = [Ag+][Br-] = (x)(0.10 + x)

Substitute the given Ksp value into the equation:

5.0×10^-13 = (x)(0.10 + x)

Now, rearrange the equation and solve using the quadratic formula as explained in the previous case.

Make sure to calculate both solutions for "x" (positive and negative) but discard the negative value since it doesn't make sense in this context.

Note: If the quadratic formula doesn't work or gives you complex solutions, it's possible that simplifications or assumptions were made in the original problem or the values provided are incorrect.