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?
Write the equation and balance it.
moles HCl = M x L = ??
Using the coefficients in the balanced equation, convert moles HCl to moles Pb(NO3)2.
a) M = moles/L
b) g = moles x molar mass.
how do i write the equation?
To solve this problem, we need to use the concept of stoichiometry and the balanced chemical equation for the reaction between HCl and Pb(NO3)2. The balanced equation is as follows:
2HCl (aq) + Pb(NO3)2 (aq) -> 2PbCl2 (s) + 2HNO3 (aq)
a) To determine the original concentration of Pb(NO3)2, we can use the concept of mole ratios or the equation's coefficients. From the balanced equation, we know that 2 moles of HCl react with 1 mole of Pb(NO3)2 to produce 2 moles of PbCl2.
Given that 56 ml of 0.022 M HCl solution was used, we can calculate the moles of HCl:
moles of HCl = volume (L) × concentration (M)
= 0.056 L × 0.022 M
= 0.001232 moles
Since the mole ratio between HCl and Pb(NO3)2 is 2:1, the moles of Pb(NO3)2 will be half of the moles of HCl:
moles of Pb(NO3)2 = 0.001232 moles / 2
= 0.000616 moles
Now, we need to determine the original concentration of Pb(NO3)2. We know that the volume of the Pb(NO3)2 solution used is 50 ml, or 0.050 L. Using the formula:
concentration (M) = moles / volume (L)
Original concentration of Pb(NO3)2 = 0.000616 moles / 0.050 L
= 0.01232 M
Therefore, the original concentration of Pb(NO3)2 is 0.01232 M.
b) To calculate the number of grams of PbCl2 that will precipitate, we need to determine the moles of PbCl2 formed in the reaction. From the balanced equation, we know that the mole ratio between Pb(NO3)2 and PbCl2 is 1:1.
Given that the moles of Pb(NO3)2 is 0.000616 moles, the moles of PbCl2 formed will also be 0.000616 moles.
To convert moles to grams, we need to use the molar mass of PbCl2, which is the sum of the atomic masses of Pb (207.2 g/mol) and Cl (35.45 g/mol), multiplied by 2 due to the subscript 2:
molar mass of PbCl2 = (207.2 + 35.45) g/mol × 2
= 284.3 g/mol
Now, we can calculate the mass of PbCl2:
mass of PbCl2 = moles × molar mass
= 0.000616 moles × 284.3 g/mol
= 0.1768 grams
So, approximately 0.1768 grams of PbCl2 will precipitate.