# What volume of .150 M potassium iodide solution will completely react with .155 L of .112 M lead (II) nitrate? equation: 2KI (aq) + Pb(NO3)2 (aq) ----> 2KNO3 (aq) + PbI2 (s)

Jun 24, 2017

$0.231$ $\text{L KI solution}$

#### Explanation:

(I'll use the balanced equation given in the question for this answer).

To solve this equation, let's first use the molarity equation to solve for the moles of lead(II) nitrate:

"moles solute" = ("molarity")("liters solution")

"mol Pb(NO"_3")"_2 = (0.112"mol"/(cancel("L")))(0.155cancel("L"))

= color(red)(0.0174 color(red)("mol Pb(NO"_3")"_2

Now, we can use the coefficients of the chemical equation to calculate the relative number of moles of potassium iodide, $\text{KI}$:

color(red)(0.0174)cancel(color(red)("mol Pb(NO"_3")"_2))((2color(white)(l)"mol KI")/(1cancel("mol Pb(NO"_3")"_2))) = color(blue)(0.0347 color(blue)("mol KI"

Now, let's use the molarity equation again to solve for the liters of the potassium iodide solution:

"liters solution" = ("moles solute")/("molarity")

"L KI solution" = (0.0347cancel("mol KI"))/(0.150(cancel("mol"))/("L")) = color(purple)(0.231 color(purple)("L KI solution"

Therefore, you would need color(purple)(0.231 sfcolor(purple)("liters" of the $0.150 M$ potassium iodide solution to completely react with the given lead(II) nitrate solution.