I'm sorry here is another I do not understand.
How do you find out exactly how much of a solution you would need to completely precipitate another compound in the reaction?
We can precipitate out solid lead(II) iodide from an aqueous lead(II) nitrate solution by adding potassium iodide. If we have 500. mL of 0.632 M lead(II) nitrate solution, how much 1.50 M potassium iodide must we add to exactly precipitate all the lead(II) ions?
2 KI(aq) + Pb(NO3)2(aq) �¨ PbI2(s) + 2 KNO3(aq)
We worked the problem last night but they wanted to know grams PbI2 pptd. In this problem, the only difference is that they want to know how much 1.50M KI must be added.
1. mols Pb(NO3)2 = M x L = ?
2. Convert mols Pb(NO3)2 to mols KI using the coefficients in the balanced equation.
3. The M KI = mols KI/L KI. You know M and mols, solve for L. Convert to mL if you wish.
To determine the amount of potassium iodide needed to completely precipitate all the lead(II) ions, you can follow these steps:
1. Write a balanced chemical equation for the reaction:
2 KI(aq) + Pb(NO3)2(aq) → PbI2(s) + 2 KNO3(aq)
2. Determine the stoichiometry of the reaction. From the balanced equation, you can see that 2 moles of potassium iodide (KI) react with 1 mole of lead(II) nitrate (Pb(NO3)2) to form 1 mole of lead(II) iodide (PbI2). This stoichiometry allows us to find the mole ratio between the two reactants.
3. Calculate the number of moles of lead(II) nitrate:
moles of Pb(NO3)2 = concentration (M) × volume (L)
moles of Pb(NO3)2 = 0.632 M × 0.500 L
4. Convert the moles of lead(II) nitrate to moles of lead(II) iodide using the stoichiometry from the balanced equation. Since the stoichiometry is 2:1, the moles of Pb(NO3)2 will be the same as the moles of PbI2 that will precipitate:
moles of PbI2 = moles of Pb(NO3)2
5. Convert the moles of lead(II) iodide to the mass of lead(II) iodide using its molar mass. The molar mass of PbI2 is the sum of the atomic masses of lead (Pb) and iodine (I), which are 207.2 g/mol and 126.9 g/mol, respectively:
mass of PbI2 = moles of PbI2 × molar mass of PbI2
6. Determine the amount of potassium iodide needed. Since the stoichiometry of the reaction is 2:1 for KI:Pb(NO3)2, the moles of potassium iodide required will be twice the moles of lead(II) nitrate:
moles of KI = 2 × moles of Pb(NO3)2
7. Calculate the volume of 1.50 M potassium iodide solution needed to obtain the required moles of potassium iodide:
volume (L) = moles of KI ÷ concentration (M)
volume (L) = moles of KI ÷ 1.50 M
Once you have performed these calculations, you will have the volume of 1.50 M potassium iodide solution needed to completely precipitate all the lead(II) ions.