Answer to Question #98117 in General Chemistry for Gia Smith

Question #98117
Solid sodium chloride is slowly added to 75.0 mL of a lead(II) nitrite solution until the concentration of chloride ion is 0.0232 M. The maximum amount of lead ion remaining in solution is _____M
1
Expert's answer
2019-11-06T07:47:49-0500

Q98117


Solution:

As solid Sodium chloride "(NaCl)" is added to the lead(II) nitrite solution, volume of the solution does not change effectively.

Thus, final moles of chloride ion "(Cl^-)=0.0232*75\/1000"

"=1.74 milimoles." (Given)


According to the chemical reaction given below:


"2NaCl(aq)+Pb(NO_2)_2(aq) \\to PbCl_2" "(s)+2NaNO_2(aq)"


2 moles of Sodium chloride reacts with 1 mole of "Pb(NO_2)_2" and 1 mole of it contains 1 mole of Pb2+ ions.

Thus, 2 moles of "Cl^-" reacts with 1 mole of "Pb^{2+}" ions.


Hence, final concentration of lead ions is

"=1.74\/2 milimoles."

"=0.87 milimoles."

"=0.87*10^{-3}\/(75\/1000)M"

"=0.0116 M." (Answer)



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