Answer:
a. V = 0.1736 L of 0.200 KCl, or 174 ml KCl to 3 sig figs.
Step-by-step explanation:
The chemical equation
2KI(aq) + Pb(NO₃)₂(aq) → 2KNO₃(aq) + PbI₂(s)
tells us we need 2 moles of KCl for every 1 mole of Pb(NO₃)₂ to produce 2 moles of KNO₃ and 1 mole of PbI₂.
Lets start by determining how many moles of Pb(NO₃)₂ are contained in 155.0 ml of a 0.112M solution of the compound. Remember the definition of M (molar). It is moles per liter. 0.112M is telling us that there are 0.112 moles of Pb(NO₃)₂ in each liter of solution. We are given 155 ml of the stuff (metric for compound), which is 0.155 Liter. Moles is the product of the Concentration and the Volume:
Moles = C*V
Moles Pb(NO₃)₂ = (0.112 moles/liter)*(0.155 liters) = 0.01736 moles Pb(NO₃)₂
The balanced equation says we need twice as many moles of KCl as we have of Pb(NO₃)₂, or:
(2 moles KCL)/(1 mole Pb(NO₃)₂)
So to completely precipitate the Pb(NO₃)₂ we'll need:
[2 moles KCL)/(1 mole Pb(NO₃)₂]*(0.01736 moles Pb(NO₃)₂
This equals 0.03472 moles of KCl
We have 0.200 M KCl. Rewrite that as 0.200 moles/liter KCl.
To find the volume required that will contain 0.03472 moles of KCl can be determined by:
[Concentration] * [Volume] = Moles
(0.200 moles KCl/liter)*(V) = 0.03472 moles of KCl
V = 0.1736 L or 174 ml to 3 sig figs
Answer option a.