Find the Concentration of CrO4^2- in 0.0033M Ba(NO3)2 saturated with BaCrO4. Include activity coefficients in your solubility-product expression. Ksp of BaCrO4 = 1.17 x 10^-10
I have no idea what to do to solve this :(
Calculate ionic strength of 0.0033M Ba(NO3)2. Then calculate activity coefficient (or some books have a table that can be used).
Then [(Ba^2+)*fBa] x [(CrO4^2-)*fCrO4] = Ksp = 1.17E-10
Substitute (Ba^2+) = 0.0033M + x
for (CrO4^2-) substitute x
Substitute fBa and fCrO4 and solve for x.
Thank you for answering! I got .0099 for the ionic strength. I just don't understand how to calculate the activity coefficient though. We just learned the concept and I'm confused.
To find the concentration of CrO4^2- in 0.0033M Ba(NO3)2 saturated with BaCrO4, we need to make use of the solubility product constant (Ksp) of BaCrO4.
The solubility product constant expression for BaCrO4 is as follows:
Ksp = [Ba^2+][CrO4^2-]
We are given the Ksp value for BaCrO4 as 1.17 x 10^-10. Now, let's assume that the solubility of BaCrO4 is "s" moles per liter. Since Ba(NO3)2 dissociates into 2 moles of Ba^2+ ions, the concentration of Ba^2+ is 2 * 0.0033 = 0.0066M.
Using the stoichiometry of the balanced chemical equation for the dissociation of BaCrO4, we can conclude that the concentration of CrO4^2- ions is also "s" moles per liter.
The activity coefficient (γ) is a measure of how much the concentration of ions deviates from the ideal behavior due to interactions between ions in solution. In this case, the activity coefficient for BaCrO4 can be neglected, as the ions are not highly charged and do not experience significant ion-ion interactions. However, we still need to consider the activity coefficient for Ba^2+ ions.
For simplicity, we will assume an activity coefficient (γ) of 1 for the Ba^2+ ions in this explanation.
Plugging in the known values into the solubility product expression:
Ksp = (0.0066 * γ)(s * γ)
Rearranging the equation, we get:
s^2 = Ksp / (0.0066 * γ)^2
Now, substituting the values:
s^2 = (1.17 x 10^-10) / (0.0066 * γ)^2
Taking the square root of both sides, we get:
s = sqrt((1.17 x 10^-10) / (0.0066 * γ)^2)
Finally, the concentration of CrO4^2- ions is equal to the solubility, s.