A student prepares a soln that is initally 0.006M Pb2+ and 0.00450M I-. This soln is placed into a 1.00cm wide cuvet and inserted into the spectrometer producing as abosrbance reading of 0.430 at a wavelength of 456nm. What is the equilibrium concentration of the iodide ions in the soln?
From a graoh they give us y=188x+0.0541 where y is absorbance and x is concentration of I-
I plugged inn .430 for y and got
.430=188x+0.0541 x=0.002MI-
Is this all the question is asking?
I think you have interpreted correctly IF (that's a big IF) the graph is reading absorbance versus concentration of I^-.
No, the question is asking for the equilibrium concentration of iodide ions in the solution. You have correctly used the graph to solve for the concentration of iodide ions (x), which you found to be 0.002 M. However, this concentration is not necessarily the equilibrium concentration.
To determine the equilibrium concentration of iodide ions, you need to consider the initial concentrations of lead (II) ions (Pb2+) and iodide ions (I-) and their reaction using the given stoichiometry.
The balanced chemical equation for the reaction between lead (II) ions and iodide ions is:
Pb2+ + 2I- -> PbI2
By using an ICE (Initial, Change, Equilibrium) table, you can determine the equilibrium concentration of iodide ions. Let's set up the table:
Initial:
[Pb2+] = 0.006 M
[I-] = 0.00450 M
Change:
[Pb2+] decreases by x M (considering that 1 mole of Pb2+ reacts with 2 moles of I-)
[I-] decreases by 2x M
Equilibrium:
[Pb2+] = 0.006 - x M
[I-] = 0.00450 - 2x M
From the balanced equation, we know that the stoichiometric ratio between Pb2+ and I- is 1:2. Since the initial concentration of Pb2+ is double that of I-, we can assume that x will be a small value compared to the initial concentrations. Therefore, we can approximate 2x as negligible compared to 0.006 M and treat [Pb2+] as constant.
Now, we can express the relationship between absorbance (y), concentration of iodide ions (x), and the equilibrium concentration of iodide ions ([I-]) as given by the graph equation:
y = 188x + 0.0541
Substituting the equilibrium concentration of iodide ions ([I-]) for x, we have:
0.430 = 188([I-]) + 0.0541
Solving this equation will give you the equilibrium concentration of iodide ions in the solution.