Question

12. (10 pts) Fluoride ions dissolved in drinking water can be detected using aluminum chloride and an indicator that reacts with the released AB ions. The net ionic equation is shown below: AIF3+3 Cl AICls+ 6 F The normal fluoride concentration in New York State tap water is 1.00 mg/L. If a 0.150 L sample of normal tap water is titrated with a 3.00 x 104 M solution of AlCl3, what volume in mL of the AlCls solution is required to completely react with these ions? (mL= 100 Volume of AlCla solution: mL

Answer 1

Given, the net ionic equation,

AlCl₃ + 6F^- $\rightarrow$ AlF₆^3- + 3Cl^-

Also given,

Volume of sample of normal tap water = 0.150 L

Fluoride concentration in tap water = 1.00 mg /L

Concentration of AlCl₃ solution = 3.00 x 10^-4 M

Now, Firstly calculating the number of moles of fluoride in 0.150 L of tap water,

= 0.150 L x (1.00 mg / 1 L) x (1g / 1000 mg) x ( 1 mol /19 g)

= 7.8947 x 10^-6 moles of fluoride

Now, Using the calculated moles of fluoride and mole ratio from the balanced chemical reaction, calcualting the number of moles of AlCl₃ required to react.

= 7.8947 x 10^-6 moles of fluoride x ( 1 mol of AlCl₃ / 6 mol of fluoride ions)

= 1.3158 x 10^-6 moles of AlCl₃ are required.

Now, using the given concentration of AlCl₃ and the calculated moles of AlCl_3, calculating the volume of AlCl₃ required,

We know,

Molarity = number of moles / L of solution

3.00 x 10^-4 M =1.3158 x 10^-6 moles of AlCl₃ / L of solution

Volume of AlCl₃ in L = 0.004386 L x ( 1000 mL / 1L)

Volume of AlCl₃ solution required = 4.38 mL Or 4.39 mL