Asked by CL

the lead (II) nitrate in 25.49 ml of a 0.1338M solution reacts with all of the aluminum sulfate in 25.00 ml solution. What is the molar concentration of the aluminum sulfate in the original aluminum sulfate solution?
Answered by Doc
The reaction is a 3 to 1 ratio Pb(NO3)2 : Al2(SO4)3 => moles aluminum sulfate = 1/3(moles lead nitrate) => 1/3[(0.02549L)(0.1338M)] = 0.00114 mole aluminum sulfate => Therefore, concentration = [Al2(SO4)3] = 0.00114 mole/0.02549L = 0.045M Aluminum Sulfate.