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

The molarity of the aluminum chloride solution is 0.441 M.

Step-by-step explanation:

To determine the molarity of the aluminum chloride solution, we can set up a stoichiometric mole ratio with the silver nitrate solution. Since the equation for the reaction between aluminum chloride and silver nitrate is:

AlCl3(aq) + 3AgNO3(aq) → Al(NO3)3(aq) + 3AgCl(s)

We can see that the mole ratio of aluminum chloride to silver nitrate is 1:3. Therefore, if 13.50 mL of the aluminum chloride solution reacts with 10.00 mL of 0.109 M silver nitrate solution, the amount of aluminum chloride is:

13.50 mL * (0.109 mol/L) * (1 L / 1000 mL) = 0.00147125 mol

Since the mole ratio is 1:3, the amount of silver nitrate is:

0.00147125 mol * (1 mol AgNO3 / 1 mol AlCl3) = 0.00441375 mol

Now, we can calculate the molarity of the aluminum chloride solution:

Molarity = (0.00441375 mol) / (10.00 mL * (1 L / 1000 mL)) = 0.441 M

Answer 2

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.

Step-by-step 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.