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?
1 answer
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.