Given, the net ionic equation,
AlCl3 + 6F- AlF63- + 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 AlCl3, 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
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