Solutions to Exercises

Exercise 1. Is hydrofluoric acid, HF, a strong, weak, or nonelectrolyte?

HF is a weak acid and only partially ionizes. It is a weak electrolyte.

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Exercise 2. Indicate each of the following as a nonelectrolyte, weak electrolyte or strong electrolyte.

a) FeCl₃ This is a soluble ionic compound. It is a strong electrolyte.

b) H₂SO₃ This is a weak acid, therefore, it is a weak electrolyte.

c) C₆H₁₂O₆ This is a sugar, therefore, it is a nonelectrolyte

d) CaClO₄ This is an ionic compound and is soluble in water. It is a strong electrolyte

e) CH₃CH₂OH This is an alcohol and is a nonelectrolyte

f) Na₂CO₃ This is an ionic compound that is soluble in water and is a strong electrolyte

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Exercise 3. What is the molar concentration of Mg⁺² ions in an aqueous solution that is 0.455 M in Mg₃(PO₄)₂. What is the molar concentration of PO₄^3-?

Write the equation for the dissociation:

\(\displaystyle Mg_3(PO_4)_2\;(s)\;^{\underrightarrow{^{H_2O}}}\;3\;Mg^{2+}(aq)\;+\;2\;PO_4^{3-}\;(aq)\)

For each mole of Mg₃(PO₄)₂ there are 3 moles of Mg⁺² ions and 2 moles of PO₄^3- ions.

3 x 0.455 M Mg⁺² = 1.365 M Mg⁺²
2 x 0.455 M PO₄^3- = 0.91 M PO₄^3-

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Exercise 4. What is the total molar concentration of ions in an aqueous 1.570 M K₂CrO₄?

Write the dissociation equation:

\(\displaystyle K_2CO_3\;(s)\;^{\underrightarrow{^{H_2O}}}\;2\;K^+(aq)\;+\;CO_3^{2-}\;(aq)\)

One mol of potassium carbonate contains 2 moles of K₊ ions and 1 mole of CO₃^2- ions.

2 x 1.570 M K₊ + 1.570 M CO₃^2- = 4.71 M ions

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Exercise 5. An aqueous solution is prepared by mixing 0.625 g of K₂SO₄, 1.264 g of K₃PO₄, and 0.543 g Na₂SO₄ and diluting with water to 250.00 mL. What is the molar concentration of each ion in the solution?

First we need to convert the grams of ionic compounds to moles. From this we can determine the molarity of each ion and finally the concentration. First calculate the molar masses of the compounds:

M_m K₂SO₄ = 174.259 g/mol
M_m K₃PO₄ = 212.27 g/mol
M_m Na₂SO₄ = 142.04 g/mol

Convert g of substances to mols

\(\displaystyle 0.625\;g\;K_2SO_4\times\frac{1\;mol\;K_2SO_4}{174.259\;g\;K_2SO_4}\;=\;\mathbf{0.0035\underline{8}66\;mol\;K_2SO_4}\)
 
\(\displaystyle 1.264\;g\;K_3PO_4\times\frac{1\;mol\;K_3PO_4}{212.27\;g\;K_3PO_4}\;=\;\mathbf{0.0059\underline{5}47\;mol\;K_3PO_4}\)
 
\(\displaystyle 0.543\;g\;Na_2SO_4\times\frac{1\;mol\;NaK_2SO_4}{142.04\;g\;Na_2SO_4}\;=\;\mathbf{0.0038\underline{2}29\;mol\;Na_2SO_4}\)
 

Next, determine the number of moles of each ion:

2 x 0.0035866 mol K⁺ + 3 x 0.0059547 mol K⁺ = 0.0250 mol K⁺

1 x 0.0059547 mol PO₄^3- = 0.0059547 mol PO₄^3-

1 x 0.0035866 mol SO₄^2- + 1 x 0.0038229 mol SO₄^2- = 0.00741 mol SO₄^2-

Finally, we calculate the molar concentration of each ion by dividing the number of moles of each ion by the volume in liters.

\(\displaystyle\frac{0.0250\;mol\;K^+}{0.25000\;L}\;=\;\mathbf{0.100\;M\;K^+}\)
 
\(\displaystyle\frac{0.005947\;mol\;PO_4^{3-}}{0.25000\;L}\;=\;\mathbf{0.0219\;M\;PO_4^{3-}}\)
 
\(\displaystyle\frac{0.00741\;mol\;SO_4^{2-}}{0.25000\;L}\;=\;\mathbf{0.0296\;M\;SO_4^{2-}}\)
 

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