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)
Comments
Leave a comment