Question

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.

Answer 1

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:

$AlCl_3(aq)+3AgNO_3(aq)->3AgCl(s)+Al(NO_3)_3(aq)$

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 :

$Molarity of Alcl_3=3 x 0.010Lx 0.109M/0.0135\\=0.242M$

Answer is : 0.242M.