Subscribe $4.99/month

Un-lock Verified Step-by-Step Experts Answers.

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

Our Website is free to use.
To help us grow, you can support our team with a Small Tip.

All the data tables that you may search for.

Need Help? We got you covered.

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Products

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Need Help? We got you covered.

Chapter 4

Q. 4.13

When hydrofluoric acid, HF(aq), gets on the skin, it migrates away from the site of exposure and shuts down the microcapillaries that carry blood. The clinical management of burns on the skin from exposure to hydrofluoric acid may include the injection of a soluble calcium salt in the skin near the site of contact to stop the migration of HF(aq). As F^{-} ions from the acid move through the skin, they combine with Ca^{2+} ions and form deposits of insoluble CaF_{2}. If 1.00 mL of a 2.24 M aqueous solution of Ca^{2+} is injected, what is the mass of CaF_{2} produced if all the calcium ions react with fluoride ions?

Step-by-Step

Verified Solution

Collect and Organize We are given the volume and concentration of a solution of Ca^{2+} ions and asked to determine the maximum amount of CaF_{2} precipitate that could be formed upon reaction with F^{-} ions.

Analyze The net ionic reaction for the process involved is

Ca^{2+}(aq) + 2 F^{-}(aq) → CaF_{2}(s)

The problem states that all the calcium ions react, so we calculate the mass of product, assuming that Ca^{2+} ions are the limiting reactant.

Solve We can calculate the number of moles of Ca^{2+}(aq) and use the stoichiometric relationships in the net ionic equation to determine the mass of product. We will also need the molar mass of CaF_{2}:

\mathscr{M}_{CaF_{2}}=40.08  g/mol +2(19.00  g/mol) = 78.08  g/mol  CaF_{2}

1.00 \times 10^{-3}  \sout{L  Ca^{2+}}\times \frac{2.24  \sout{mol  Ca^{2+}}}{1  \sout{L  Ca^{2+}}} \times \frac{1  \sout{mol  CaF_{2}}}{1  \sout{mol  Ca^{2+}}} \times \frac{78.08  g   CaF_{2}}{1  \sout{mol  Ca^{2+}}}=0.175  g  CaF_{2}

Think About It This problem asks for the mass of CaF_{2} that could be formed, assuming that Ca^{2+} ions were the limiting reactant. However, when this treatment is actually used, it is more likely that the Ca^{2+} ions will be in excess to make sure all the fluoride ions are consumed.