A titration was performed in which 56 ml of a .022M HCL solution was used to titrate 50 ml of a PB(NO3)2 solution

a) what was original concentration of the Pb(NO3)2?
b) How many grams of PBCl2 will ppt?

To find the original concentration of the Pb(NO3)2 solution and determine how many grams of PbCl2 will precipitate, we need to use the concept of stoichiometry and the balanced chemical equation for the reaction between HCl and Pb(NO3)2.

The balanced chemical equation for the reaction is:
2HCl + Pb(NO3)2 -> 2HNO3 + PbCl2

a) To determine the original concentration of Pb(NO3)2, we need to use the concept of stoichiometry and the volume and concentration of HCl used in the titration. Here's how to calculate it:

1. Write down the balanced chemical equation.
2HCl + Pb(NO3)2 -> 2HNO3 + PbCl2

2. Determine the mole ratio between HCl and Pb(NO3)2.
From the balanced equation, we can see that there is a 2:1 mole ratio between HCl and Pb(NO3)2.

3. Calculate the moles of HCl used.
Moles of HCl = volume of HCl (in L) × concentration of HCl (in mol/L)
Moles of HCl = 0.056 L × 0.022 mol/L = 0.001232 mol

4. Use the mole ratio to find the moles of Pb(NO3)2.
Moles of Pb(NO3)2 = Moles of HCl ÷ 2
Moles of Pb(NO3)2 = 0.001232 mol ÷ 2 = 0.000616 mol

5. Calculate the original concentration of Pb(NO3)2.
Original concentration of Pb(NO3)2 = Moles of Pb(NO3)2 ÷ volume of Pb(NO3)2 (in L)
Original concentration of Pb(NO3)2 = 0.000616 mol ÷ 0.050 L = 0.01232 M

Therefore, the original concentration of Pb(NO3)2 is 0.01232 M.

b) To determine how many grams of PbCl2 will precipitate, we need to use the moles of Pb(NO3)2 and the molar mass of PbCl2.

1. Calculate the moles of PbCl2 using the moles of Pb(NO3)2.
Moles of PbCl2 = Moles of Pb(NO3)2

2. Determine the molar mass of PbCl2.
The molar mass of PbCl2 can be calculated by summing the atomic masses of each element in the compound: Pb (207.2 g/mol) + 2Cl (2 * 35.45 g/mol) = 278.1 g/mol

3. Calculate the mass of PbCl2.
Mass of PbCl2 = Moles of PbCl2 × molar mass of PbCl2
Mass of PbCl2 = 0.000616 mol × 278.1 g/mol = 0.171 g

Therefore, approximately 0.171 grams of PbCl2 will precipitate.