Answer:
If 13.50 mL of an aluminum chloride solution is needed to reach the equivalence point with 10.00 mL of 0.109 M silver nitrate solution, determine the molarity of the aluminum chloride solution.
Explanation:
The balanced chemical equation of the reaction is:
[tex]AlCl_3(aq)+3AgNO_3(aq)->3AgCl(s)+Al(NO_3)_3(aq)[/tex]
So, one mole of aluminum chloride reacts with three moles of silver nitrate.
At the equivalence point,
the number of moles of each reactant must be equal.
The number of moles = molarity x volume in L.
Number of moles of AlCl3 = volume x molarity
=0.0135Lx Molarity
The number of moles of AgNO3 = 3 x 0.010Lx 0.109M
Thus,
0.0135Lx Molarity = 3 x 0.010Lx 0.109M
Molarity of AlCl3 :
[tex]Molarity of Alcl_3=3 x 0.010Lx 0.109M/0.0135\\=0.242M[/tex]
Answer is : 0.242M.
