Solutions: Chemistry

Question

Mixing 45.0 mL of an NaF solution with excess Pb(NO3)2 results in the precipitation of 1.29 g of PbF2 (see equation below).



2 NaF(aq) + Pb(NO3)2(aq) → PbF2(s) + 2 NaNO3(aq) A. What is the concentration (in M) of the NaF solution?

concentration (M): _______________

Write the net ionic equation for this precipitation reaction. Include the states in your equation for full credit.

equation: ____________________________________________________________

What are the spectator ions for this precipitation reaction? ______________ and ______________

Answer 1

Hello Mahala,

first change grams of PbF2 into moles— 1.29 gmsPbF2×(1mole PbF2/245.2 gm PbF2)=.00526moles PbF2

from the balanced equation you see, one mole of PbF2 is formed from two moles of NaF, so moles of NaF used in this reaction— .00526×2=.0105 moles

Conc of NaF in molarity M=.0105/.045 L=0.234 M

The net ionic equation is. Pb++(aq)+2F-(aq)=PbF2(s)

The spectator ions for this reaction are Na+ and NO3- ions

Pl contact me if you need further explanations.