50.00 mL samples of 0.200 mol/L potassium phosphate with an excess of 0.120 mol/L lead II nitrate solution.
what is the minimum volume of lead II nitrate solution required? What volume of lead II nitrate solution should the instructor tell the students to use? Describe how the students can test for completeness of reaction of the limiting reagent?
To determine the minimum volume of lead II nitrate solution required, we need to find the limiting reactant in the reaction between potassium phosphate and lead II nitrate.
1. Calculate the moles of potassium phosphate:
Moles of potassium phosphate = concentration (mol/L) × volume (L)
Moles of potassium phosphate = 0.200 mol/L × 0.050 L
Moles of potassium phosphate = 0.010 mol
2. Calculate the moles of lead II nitrate:
Moles of lead II nitrate = concentration (mol/L) × volume (L)
Moles of lead II nitrate = 0.120 mol/L × volume (L)
Since the stoichiometric ratio between potassium phosphate and lead II nitrate is 1:2, the required moles of lead II nitrate is twice the moles of potassium phosphate. Therefore:
0.010 mol (potassium phosphate) × 2 = 0.020 mol (lead II nitrate)
3. Calculate the required volume of lead II nitrate:
Volume of lead II nitrate = moles / concentration
Volume of lead II nitrate = 0.020 mol / 0.120 mol/L
Volume of lead II nitrate = 0.167 L or 167 mL
So, the minimum volume of lead II nitrate solution required is 167 mL.
To ensure complete reaction of the limiting reagent (lead II nitrate in this case), the instructor should tell the students to use a volume slightly larger than the minimum required.
One way for students to test for the completeness of the reaction is to add a few drops of a reactive substance that can detect the presence of either potassium phosphate or lead II nitrate. For example, they can add a few drops of sodium sulfate solution. If there is still a reaction occurring, it means one of the reactants is still present. Otherwise, if there is no further reaction, then it can be concluded that the reaction is complete and all the reactants have been consumed.
To determine the minimum volume of lead II nitrate solution required, we need to identify the limiting reagent in the reaction. The limiting reagent is the reactant that will be completely consumed and determines the maximum amount of product that can be formed. This can be determined by comparing the stoichiometry of the reactants.
The balanced chemical equation for the reaction between potassium phosphate (K3PO4) and lead II nitrate (Pb(NO3)2) is:
2 K3PO4 + 3 Pb(NO3)2 → 6 KNO3 + Pb3(PO4)2
From the equation, we can see that the molar ratio between K3PO4 and Pb(NO3)2 is 2:3.
Given that we have a 0.200 mol/L potassium phosphate solution and a 0.120 mol/L lead II nitrate solution, we can use the molar concentration and the volume to calculate the moles of each reactant:
moles of K3PO4 = concentration of K3PO4 × volume of K3PO4
= 0.200 mol/L × 0.050 L
= 0.010 mol
moles of Pb(NO3)2 = concentration of Pb(NO3)2 × volume of Pb(NO3)2
= 0.120 mol/L × volume of Pb(NO3)2
To determine the minimum volume of Pb(NO3)2 solution required, we set up the stoichiometric ratio:
0.010 mol K3PO4 × (3 mol Pb(NO3)2 / 2 mol K3PO4) = 0.120 mol/L × volume of Pb(NO3)2
0.015 mol Pb(NO3)2 = 0.120 mol/L × volume of Pb(NO3)2
Therefore, the minimum volume of Pb(NO3)2 solution required is:
volume of Pb(NO3)2 = 0.015 mol / 0.120 mol/L
= 0.125 L = 125 mL
This is the minimum volume needed to ensure complete reaction between the reactants.
To determine the volume of Pb(NO3)2 solution that the instructor should tell the students to use, it is common practice to add an excess of the reactant. This ensures that all of the limiting reagent is consumed and prevents incomplete reactions. In this case, the instructor should instruct the students to use a volume greater than or equal to 125 mL to ensure an excess of Pb(NO3)2.
To test for the completeness of the reaction, the students can use a simple precipitation test. They can add a few drops of a phosphate ion indicator, such as ammonium molybdate, to a small sample of the reaction mixture. If the reaction is complete, a yellow precipitate of lead phosphate (Pb3(PO4)2) will form. The absence or presence of the precipitate will indicate if the reaction has gone to completion or not.