Calculate the molar solubility of AgBr in 0.15M NaBr at 25 degree celcius (Ksp of AgBr= 5.0*10¤-3 mol¤2dm¤-6 at 25 degree celcius)?
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To calculate the molar solubility of AgBr in 0.15M NaBr at 25 degrees Celsius, we first need to determine the effect of the NaBr concentration on the solubility of AgBr.
AgBr dissociates in water according to the following equation:
AgBr(s) ↔ Ag+(aq) + Br-(aq)
The solubility equilibrium expression for AgBr can be written as:
Ksp = [Ag+][Br-]
Since we are dealing with a mixture containing both AgBr and NaBr, the presence of NaBr will affect the concentration of Ag+ and Br- in the solution.
Let's assume the molar solubility of AgBr in the presence of NaBr is x. In a saturated solution of AgBr, the concentration of Ag+ will also be x, and the concentration of Br- will be (0.15 + x) because NaBr dissociates to provide additional Br- ions.
Substituting these concentrations into the solubility product expression, we have:
Ksp = x * (0.15 + x)
The given value for Ksp is 5.0 * 10^(-3) mol^2 dm^(-6).
Now, we can solve for x:
5.0 * 10^(-3) = x * (0.15 + x)
Rearranging the equation and setting it equal to zero:
x^2 + 0.15x - 5.0 * 10^(-3) = 0
This is a quadratic equation, and we can solve it using the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / (2a)
a = 1
b = 0.15
c = -5.0 * 10^(-3)
Plugging in these values, we have:
x = (-0.15 ± sqrt((0.15)^2 - 4(1)(-5.0 * 10^(-3)))) / (2(1))
x = (-0.15 ± sqrt(0.0225 + 0.02)) / 2
x = (-0.15 ± sqrt(0.0425)) / 2
Taking the positive root, we have:
x = (-0.15 + sqrt(0.0425)) / 2
x = (-0.15 + 0.2062) / 2
x = 0.0562 / 2
x = 0.0281 mol dm^(-3)
Therefore, the molar solubility of AgBr in 0.15M NaBr at 25 degrees Celsius is 0.0281 mol dm^(-3).
To calculate the molar solubility of AgBr in 0.15M NaBr, we need to use the concept of common ion effect.
The common ion effect states that the solubility of a slightly soluble salt is reduced when a common ion is present in the solution. In this case, NaBr is a salt containing the common ion "Br-".
Given:
Ksp of AgBr = 5.0 * 10^-3 mol^2 dm^-6
NaBr concentration = 0.15M
Step 1: Write the balanced chemical equation of the dissolution of AgBr in water:
AgBr(s) ⇌ Ag+(aq) + Br-(aq)
Step 2: Write the expression for the solubility product constant (Ksp) using the concentration of Ag+ and Br- ions:
Ksp = [Ag+][Br-]
Step 3: Set up the equilibrium expression for AgBr dissolving in water:
Ksp = [Ag+][Br-] = x * (0.15 + x)
Here, "x" represents the molar solubility of AgBr in moles per liter (mol/L) when it dissolves.
Step 4: Since the NaBr dissociates completely, the concentration of Br- in the solution is (0.15 + x) M.
Step 5: Substitute the given values and solve for x:
Ksp = x * (0.15 + x)
5.0 * 10^-3 = x * (0.15 + x)
Since Ksp is very small, we can assume that the value of "x" is much smaller compared to 0.15. Therefore, we can ignore "x" in the (0.15 + x) term.
5.0 * 10^-3 = x * 0.15
Solving for x:
x = (5.0 * 10^-3) / 0.15
x ≈ 3.33 * 10^-4 mol/L
So, the molar solubility of AgBr in 0.15M NaBr at 25 degrees Celsius is approximately 3.33 * 10^-4 mol/L.