A student pipets 10.00 mL of a 0.100 M solution of reagent A into an Erienmeyer
ID: 482848 • Letter: A
Question
A student pipets 10.00 mL of a 0.100 M solution of reagent A into an Erienmeyer flask. Reagent A is then titrated with a titrating solution containing reagent B. The following reaction occurs A + 3B Products It takes the addition of 33.24 mL of the titrating solution (i.e., the reagent B solution) to reach the endpoint. What is the molar concentration of reagent B in the titrating solution? Assume that a student places 1.0 mol of solid Pbl_2 in 1.0 L of water. Part of the lead iodide dissolves according to the following reaction Pl_2 (s) doubleheadarrow Pb^2+ + 2I^- (aq) After allowing the solution to sit for 30 minutes the student determines that the iodide concentration was [I] = 0.0032 M. What is the Pb^2+ concentration? Assume that a student places 1.0 mol of solid PbBr, in 1.0 L of 0.0010 M Pb^2+ Part of the lead bromide dissolves according to the following reaction PbBr_2 (s) doubleheadarrow Pb^2+ (aq) + 2Br_(aq) After allowing_ the solution to sit for 30 minutes the student determines that the bromide concentration was [Br] = 0.0012 M What is the Pb^2+ concentration?Explanation / Answer
2) As per the given equation ,
1mole of A will react with 3 moles of B
the moles of A added = Molarity x volume = 0.1M X 10mL = 1 millimoles
moles of B needed = 3 X moles of A added = 3 X 1 millimoles = 3 millimoles
Moles = Molarity X volume
Molarity of B = Moles of B needed / Volume in mL = 3 millimoles / 33.24 = 0.09 molar
3) as per the given equation
one mole of PbI2 will dissociate into one mole of Pb+2 and two moles of [I-]
So if concentration of [I-] = 0.0032 M
The concentration of Pb+2 = 0.5 X [I-] = 0.0016 M
4) As above one mole of PbBr2 will give one mole of Pb+2 ions and two moles of Br- ions
so if [Br-] = 0.0012 M
[Pb+2] = 0.5 X [Br-] = 0.0012 / 2 = 0.0006 M
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.