Question

How would you prepare 250 mL of a 0.100 M solution of fluoride ions from solid CaF2?

Answer 1

In 250 mL of volumetric flask add 0.975875 grams of $CaF_2$ and dissolve it in the 250 mL of water.

Given:

The solid of calcium fluoride.

To prepare:

The 250 mL solution of 0.100 M of fluoride ions from solid calcium fluoride.

Method:

Molarity of the fluoride ion solution needed = M = 0.100 M

The volume of the fluoride ion solution needed = V = 250 mL

$1 mL = 0.001L\\V=250 mL=250\times 0.001 L=0.250 L$

The moles of fluoride ion needed = n

According to the definition of molarity:

$M=\frac{n}{V}\\0.100M=\frac{n}{0.250 L}\\n=0.100M\times 0.250 L=0.025 mol$

Moles of fluoride ion = 0.025 mol

We know that solid calcium fluoride dissolves in water to give calcium ions and fluoride ions.

$CaF_2(s)\rightarrow Ca^{2+}(aq)+2F^-(aq)$

According to reaction, 2 moles of fluoride ions are obtained from 1 mole of calcium fluoride, then 0.025 moles of fluoride ions will be obtained from:

$=\frac{1}{2}\times 0.025 mol=0.0125 \text{mol of } CaF_2$

Moles of calcium fluoride = 0.0125 mol

Mass of calcium fluoride needed to prepare the solution :

$=0.0125 mol\times 78.07 g/mol=0.975875 g$

Preparation:

• Weight 0.975875 grams of calcium fluoride
• Add weighed calcium fluoride to a volumetric flask of the labeled volume of 250 mL.
• Now add a small amount of water to dissolve the calcium fluoride completely.
• After this add more water up to the mark of the volumetric flask of volume 250 mL.

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