Question

13 Determination of Ke for an Equilibrium System Pre-laboratory Assignment Section 1. Calculate [Fe Jti for test tubes 1-3. Show your work here for one of the tubes. 2. A student performed a lab similar to this one in an attempt to find Kny for the equilibrium X"(ap + Ytap g ga In one trial. the student combined 5.00 mL of system A 0.10 M X with 10.00 mL of 0.20 M Y in a test tube. When the resulting solution was analyzed with a spectrophotometer set at -600 nm, 4.00% of the light passed through the solution. For this equilibrium system, the Beer's law constant, B, is known to be 92.4 L.mol Zni A. Calculate the initial molarity of each reactant added to the test tube at the moment of mixing, [X' Jan and [YJan. (HINT: these are dilution calculations.) B. Calculate the value of A, then find [ZJequil. C. Set up an I.C.E. table to find [X"Jequil and [Y Joquit D. Find the value of Ke for this equilibrium system. evised 4/6/2016 DN

Answer 1

2.A. initial concentration of X⁺ = 0.10M

Initial concentration of Y^- = 0.20M

Volume of X⁺ added= 5.00 mL

Volume of Y^- added = 10.00mL

Total volume =15.00mL

So, the molarity of X⁺ at the moment of mixing = 5.00mL×0.10M÷15.00mL = 0.033M

Molarity of Y ^- at the moment of mixing = 10.00mL×0.20M÷15.00mL =0.133M

B) Given , the 4.00% of light passed through the solution i.e., the solution transmitted 4.00% of light. Hence , it must have absorbed 96.00% of light.

Absorbance (A)=0.96

  

=>0.96=92.4 L mol^-1 × C

=>C =(0.96÷92.4) mol L^-1 = 0.01M

So [Z]_equil. = 0.01M

C) X⁺ (aq) + Y^- (aq)    $\rightleftharpoons$   Z(aq)

Initial 0.03 0.13 0

At equil.0.03-x 0.13-x. x

In problem B we got [Z]_equil. = 0.01M

So [X⁺]_equil = 0.03-0.01 = 0.02M

& [Y^-]_equil = 0.13-0.01= 0.12M

D) equilibrium constant ,K _c is given by

K _c= [Z]_equil / [X⁺]_equil [Y^-]_equil

  = 0.01M/ (0.02M×0.12M)

= 4.17 L mol^-1