Quantitative anaylsis of cl- ion is often performed by a titration with silver nitrate, using sodium chromate as an indicator. As standardized AgNo3 is addded, both white AgCl and red Ag2CrO4 precipitate, but so long as some cl- remains, the Ag2CrO4 dissovles as the mixgture is stirred, When the red color is permanent, the equivalence point has been reached.
a) calculate the equilibrium constant for the reaction:
2AgCl +CrO4=Ag2CrO4 +2Cl-
b)if 36.00cm^3 of 0.2200 m NaCl is mixed with 36.00cm^3 of 0.2200 m AgNO3, what is the concentration of Ag remaining in the solution?
For a), use Hess' Law.
AgCl(s) ==> Ag^+ + Cl^-
Ag2CrO4(s) ==> 2Ag^+ + CrO4^-2
b)Since these are equal amounts of NaCl and AgNO3, this will be a saturated solution of AgCl(s) and the solubility is governed by the Ksp for AgCl.
AgCl ==> Ag^+ + Cl^-
Ksp = (Ag^+)(Cl^-)
Set up an ICE chart and solve for Ag^+/
Post your work if you get stuck.
i don't understand how and why you use Hess's law to find the equilbiruim constant
I will number the equations.
1) AgCl --> Ag^+ + Cl^- Ksp = ??
2)Ag2CrO4 ==> 2Ag^+ + CrO4^-2 Ksp = ??
Technically, I may not have named it right. Perhaps I should have said in a Hess' Law process.
Multilply equation 1 by 2 (which will make Ksp for AgCl squared.
Reverse equation 2 (which will mean the reciprocal of Ksp for Ag2CrO4. Then add the resulting two equations to obtain the equation the questions asks for, then Keq (what the problem wants) will be K^2(AgCl)*(1/K(Ag2CrO4).
i got part a right but i tried part b and i got an answer of 1.34 x 10^-5 but it was wrong
a) To calculate the equilibrium constant for the reaction, you need to use the balanced chemical equation:
2AgCl + CrO4^2- = Ag2CrO4 + 2Cl-
The equilibrium constant expression for this reaction is:
Kc = [Ag2CrO4][Cl-]^2 / [AgCl]^2[CrO4^2-]
We can determine the equilibrium constant by using the concentrations at a specific point during the titration. At the equivalence point, the concentration of AgCl is zero since all of it has reacted. Therefore, the equilibrium constant expression simplifies to:
Kc = [Ag2CrO4][Cl-]^2
b) To calculate the concentration of Ag remaining in the solution, we need to use the stoichiometry of the reaction and the volumes and concentrations of the reactants.
Given:
Volume of NaCl solution (V1) = 36.00 cm^3
Concentration of NaCl solution (C1) = 0.2200 M
Volume of AgNO3 solution (V2) = 36.00 cm^3
Concentration of AgNO3 solution (C2) = 0.2200 M
First, we need to determine the number of moles of Cl- present in the NaCl solution. Using the formula:
moles of Cl- = concentration of NaCl × volume of NaCl solution
moles of Cl- = 0.2200 M × 0.03600 L
moles of Cl- = 0.00792 mol
Since the stoichiometry of the reaction is 2 moles of Cl- : 1 mole of Ag, the number of moles of Ag consumed is half the number of moles of Cl-:
moles of Ag consumed = 0.00792 mol / 2
moles of Ag consumed = 0.00396 mol
Now, we need to determine the final volume of the solution after mixing:
Final volume (Vf) = V1 + V2
Final volume (Vf) = 36.00 cm^3 + 36.00 cm^3
Final volume (Vf) = 72.00 cm^3
Next, we can calculate the concentration of Ag remaining in the solution using the formula:
concentration of Ag remaining = moles of Ag remaining / final volume
moles of Ag remaining = moles of Ag initially - moles of Ag consumed
moles of Ag remaining = (C2 × V2) - moles of Ag consumed
moles of Ag remaining = (0.2200 M × 0.03600 L) - 0.00396 mol
moles of Ag remaining = 0.00792 mol - 0.00396 mol
moles of Ag remaining = 0.00396 mol
concentration of Ag remaining = 0.00396 mol / 0.07200 L
concentration of Ag remaining = 0.055 M
Therefore, the concentration of Ag remaining in the solution is 0.055 M.