. Calculation Guide and Reminder Test tubes 1-4. In these tubes, two solutions w

Question

. Calculation Guide and Reminder Test tubes 1-4. In these tubes, two solutions were mixed together, a reaction took place to form a precipitate, and a solution with different concentrations remained. It is best to think of these in terms of an i/Alfchart. 1) Part A Lines 1-5 are measured in lab. 2) Part A Line 6 is obtained using the Absorbance and the Calibration Curve provided in class. Part B 1) Obtain from molarity and volume of Pb(NO3)2. 2) Obtain from molarity and volume of KI. 3) Obtain from molarity (Line A-6) and volume at equilibrium. 4) Given initial and final, how much I reacted? 5) If X moles of I reacted, how many moles of Pb2t reacted? 6) If X moles of Pb2 reacted, how many remain at equilibrium? 7) IfX moles of Pb?* remain at equilibrium...? Test tube 5 In this tube, a precipitate dissolves so the relative concentrations are dictated by the reaction stoichiometry. An iAif chart is not really needed here. The concentration of Pb2* can be calculated directly from the concentration ofI.

Explanation / Answer

Part B :- Calculation

(1) Initial moles of Pb+2 :-

Given ;- Molarity of Pb(NO3)2=0.012 M

Volume of Pb(NO3)2 = 5.00ml = 0.005L

Moles of solute (Pb+2)= Molarity * Volume of Pb(NO3)2 = 0.012*0.005 = 0.00006 moles

(2) Initial moles of I- :-

Given :- Molarity of KI = 0.030 M

Volume of KI = 2.00 ml = 0.002 L

(1) Moles of solute (I-) = Molarity*Volume of KI = 0.03*0.002 = 0.00006 moles

(2) Moles of solute (I-) = Molarity*Volume of KI = 0.03*0.003 = 0.00009 moles

(3) Moles of solute (I-) = Molarity*Volume of KI = 0.03*0.004 = 0.00012 moles

(4) Moles of solute (I-) = Molarity*Volume of KI = 0.03*0.005 = 0.00015 moles

(5) Moles of solute (I-) = Molarity*Volume of KI = 0.03*0.005 = 0.00015 moles

(3) Equillibrium moles of I- :-

Data from Line A-6 is missing. Hence, equillibrium moles of I- can't be calculated.

Equillibrium moles of I- = Molarity*Volume at equillibrium (From line A-6)

(4) Moles of I- solid :-

Given :- Initial moles of KI = Molarity*Volume = 0.03*0.010= 0.0003 moles

(1) Final moles of KI = 0.00006 moles

I- reacted = Initial moles - Final moles = 0.0003-0.00006 = 0.00024 moles

(2) Final moles of KI = 0.00009 moles

I- reacted = Initial moles - Final moles = 0.0003-0.00009 = 0.00021 moles

(3) Final moles of KI = 0.00012 moles

I- reacted = Initial moles - Final moles = 0.0003-0.00012 = 0.00018 moles

(4) Final moles of KI = 0.00015 moles

I- reacted = Initial moles - Final moles = 0.0003-0.00015 = 0.00015 moles

(5) Final moles of KI = 0.00015 moles

I- reacted = Initial moles - Final moles = 0.0003-0.00015 = 0.00015 moles

(5) Moles of Pb+2 :-

Moles of I- reacted = 0.00024 moles

Moles of Pb+2 reacted = 0.00024 moles

(6) Equillibrium moles of Pb+2

Line A-6 data is needed to calculate equillibrium moles of Pb+2.

(7) Pb+2 at equillibrium

Line A-6 data is needed.

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