A beaker contains a solution that is 0.50 M in Au+3 ion and 0.25 M in Ag+ ion. A soultion of sodium chloride is slowly dripped into the beaker with stirring. Which ion will precipitate first, the gold or the silver? Clearly support your answer and write out your final conclusion in an explanatory sentence.
Ksp for AuCl3 = 3.2 x 10^-25
Ksp for AgCl = 1.8 × 10^-10
I'll give you some hints but it is up to you to put all of it together to make a coherent answer.
For AuCl3 ==> Ag^3+ + 3Cl^-
Write out Ksp expression and solve for (Cl^-). Plug in Ksp, (Au^3+) = 0.5M and calculate (Cl^-)
Do the same for AgCl
AgCl --> Ag^+ + Cl^-
Write out the Ksp expression, plug in Ksp and 0.25 for (Ag^+) and solve for (Cl^-).
So if you have those ions in solution and you need (Cl^-) to be what you have calculated above. which will ppt first? Of course the one that requires the smallest Cl^-. Identify that and put all of the equations and math in good order and you'll get an A+ on the assignment.
Well, it seems like these ions are having a salty party in the beaker! Let's see who gets the spotlight first.
To determine which ion will precipitate first, we need to compare their solubility product constants (Ksp). The lower the Ksp value, the less soluble the compound is in water.
The Ksp for AuCl3 is 3.2 x 10^-25, while the Ksp for AgCl is 1.8 x 10^-10. Wow, those are some tiny values!
Since the Ksp value for AuCl3 is significantly lower than that of AgCl, it means that gold chloride is way less soluble than silver chloride. So, gold chloride will precipitate first!
In conclusion, gold chloride (AuCl3) will be the first to form a solid precipitate in the beaker, followed by silver chloride (AgCl). Looks like the gold is going for the applause!
To determine which ion precipitates first, we can compare the solubility product constants (Ksp) for the chloride salts of gold and silver.
The Ksp for AuCl3 is 3.2 x 10^-25.
The Ksp for AgCl is 1.8 x 10^-10.
The solubility product constant represents the equilibrium between a solid salt and its dissociated ions in a saturated solution. The smaller the Ksp value, the less soluble the salt is and the more likely it is to precipitate.
Comparing the two Ksp values, we can see that the Ksp for AuCl3 is much smaller than the Ksp for AgCl. Therefore, gold chloride (AuCl3) is less soluble and more likely to precipitate first compared to silver chloride (AgCl).
In conclusion, the gold ion (Au+3) will precipitate first when sodium chloride is added to the beaker due to its smaller solubility product constant (Ksp) compared to silver chloride (AgCl).
To determine which ion will precipitate first, we need to compare the solubility products (Ksp) of the two chlorides, AuCl3 and AgCl.
The solubility product, Ksp, represents the maximum amount of solute that can dissolve in a solvent at a given temperature. It is calculated by multiplying the concentrations of the ions in the equilibrium state.
First, let's write the balanced equations for the formation of precipitates:
Au+3 + 3Cl- -> AuCl3
Ag+ + Cl- -> AgCl
Now, we can write the expression for Ksp for each of the precipitates:
For AuCl3: Ksp = [Au+3][Cl-]^3
For AgCl: Ksp = [Ag+][Cl-]
Given that the concentration of Au+3 is 0.50 M and the concentration of Ag+ is 0.25 M, we can calculate the initial Qsp (the reaction quotient) for each salt:
For AuCl3: Qsp = (0.50 M)(0.25 M)^3 = 0.00391
For AgCl: Qsp = (0.25 M)(0.25 M) = 0.0625
Comparing the Qsp values with the Ksp values:
For AuCl3: Qsp (0.00391) < Ksp (3.2 x 10^-25)
For AgCl: Qsp (0.0625) > Ksp (1.8 × 10^-10)
Since Qsp < Ksp for AuCl3 and Qsp > Ksp for AgCl, it means that AgCl has reached its maximum solubility limit, while AuCl3 has not.
Therefore, silver chloride (AgCl) will precipitate first from the solution before gold chloride (AuCl3).
In conclusion, when sodium chloride is added to the beaker, silver ions will precipitate as silver chloride (AgCl) before gold ions precipitate as gold chloride (AuCl3) due to the difference in their respective solubility products.