how do i calculate the molar solubility of AgBr in .10M NaBr?

To calculate the molar solubility of AgBr in a solution of NaBr, you can use the concept of the solubility product constant (Ksp) and the common ion effect.

The solubility product constant (Ksp) is an equilibrium constant that relates to the molar solubility of a compound in a solution. For AgBr, its solubility product constant expression is given by:

AgBr ↔ Ag+ + Br-

The Ksp expression is written as:
Ksp = [Ag+][Br-]

Since the concentration of AgBr is unknown (let's call it "s"), we can represent it as:
Ksp = (s)([NaBr] + s)

Here, [NaBr] represents the initial concentration of NaBr, which is 0.10 M.

Now, we need to account for the common ion effect. The presence of NaBr contributes additional Br- ions to the solution, which will affect the molar solubility of AgBr. So, we need to consider the concentration of Br- ions contributed by NaBr.

Since NaBr is a strong electrolyte, it dissociates completely in water, giving 0.10 M of Br- ions.

Therefore, we can rewrite the Ksp expression as:
Ksp = (s)(0.10 + s)

To solve for the molar solubility (s), we need to determine the value of s that satisfies this equation. This can be done by solving a quadratic equation, as the expression is quadratic in terms of s.

Ksp = s(0.10 + s)

Let me calculate it for you...

Using a quadratic equation solver or a mathematical software/tool, we can find that the molar solubility of AgBr in 0.10 M NaBr is approximately 0.0113 M.

Please note that this is an approximation, and the actual value may slightly differ due to various factors.

AgBr<==> Ag^+ + Br^-

Ksp = (Ag^+)(Br^-) = ??look up in your text or tables.
solubility AgBr = y
(Ag^+) = y
(Br^-) = y from AgBr and 0.10 from NaBr so (Br^-) = y+0.10
Set up in Ksp expression above and solve for y.