Answer to Question #194889 in General Chemistry for Prince F. Kollie

Question #194889

You mix 265.0 mL of 1.20 M lead (II) Nitrate with 300.0 mL of 1.55 M Potassium Iodide. The lead (II) Iodide is insoluble. Calculate (a)the final concentration of Pb2+ . (b) Mass of lead (II) Iodide (c) The final concentration of K+ (d) the final concentration of NO3-.


1
Expert's answer
2021-05-24T06:30:43-0400

The equation for the reaction is;

"Pb(NO_3)_{2{(aq)}}+2KI_{(aq)}\\to PbI_{2{(s)}}+2KNO_{3{(aq)}}"

a) "[Pb^{2+}]=" "0.30.L\\times 1.55M\\over0.265.L\\times 1.20M" "=1.462mol.L"

b) "Mass=n\\times M"

"n=1.86moles"

"M=461g\/mol"

"Mass(g)=1.86Moles\\times 461g\/mol=870.48g"

c) "[K^+]=" "(0.35.L\\times1.86mol)\\times 2\n=1.302mol\/L"

d) "[NO^-_3]=" "0.265.L\\times 1.20M\\over0.30.L\\times1.55M" "=0.683mol.L"

.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS