PLEASE ANSWER F-K Answers for A.5.19 B.22.6 C.13.08 D.0.4365 E.1.088 PLEASE ANSW

Question

PLEASE ANSWER F-K Answers for A.5.19 B.22.6 C.13.08 D.0.4365 E.1.088 PLEASE ANSWER E-K Pre-Laboratory Assignment for the Solublity Product Constant sp Experiment Before you can startthis experiment you must: a. Solve the problem below b. Pass an 8 question on-screen computer quiz For this pre-lab, consider the equilibrium to be Pb+2 21 Pbl You mix the following: Volume of Lead Nitrate solution 1.33 mL Original concentration of Lead Nitrate solution 0.00390 M Volume of lodide ion solution 5.75 mL, Original concentration of lodide ion solution 0.00393 M Volume of distilled water 600 mL, You measure the %T of the blank and sample at 525 nm. Assume Beer's Law, absorbance is proportional to concentration. %T of blank %T of sample 37.0% Absorbance of a 1.000 MilliMolar lodide solution Note: This constant is only for this prelab! In the experiment you will use a standard plot Calculate [to 3 decimal places, except c)] a) umol (micromoles) of Pb(II)originally put in solution Ramol b) umol of originally put in solution c) Total volume of solution d) Absorbance of sample e) M molarity of I at equilibrium umol of IT in solution at equilibrium g) umol of precipitated h) umol of Pb(Il precipitated umol of Pb(II) in solution at equilibrium j) Millimolarity of Pb(I) in solution at equilibrium k) Calculated K x10 12

Explanation / Answer

Answer:

f = 14.231

This answer can get by using formula :

( e in M ) ( c in L ) / 10-7

g = 8.369

This answer can get by using formula :

b - f = 22.6-14.231

h =8.369

This answer can get by using formula :

g/2 = 8.369 /2 = 4.1845

i = 1.005

This answer can get by using formula :

a - h = 5.19 - 4.185

j = 0.077

This answer can get by using formula :

(i / c )

( 1.005 / 13.08 ) = 0.0768

k = 91.148

This answer can get by using formula : (J)(e)2 x 103 = (0.0768)(1.088)2 (103)

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