Solutions/Answers to Exercises

Exercise 1. A 50.00 mL aqueous HCl solution is titrated with NaOH. It takes 37.62 mL of 0.254 M NaOH to reach the endpoint. What is the concentration of HCl?

The balanced chemical equation is:

HCl (aq) + NaOH (aq) → NaCl (aq) + H₂O (l)

\(\displaystyle moles\;NaOH\;=\;0.03762\;L\;\times\;0.254\;M\;=\;\mathbf{0.00956\;mol\;NaOH}\)
 
There is a 1:1 mole ratio of acid:base. Calculate the moles of HCl. At the equivalence point, all of the acid will have reacted with the base.

\(\displaystyle 0.00956\;mol\;NaOH\times\frac{1\;mol\;HCl}{1\;mol\;NaOH}\;=\;{0.00956\;mol\;HCl}\)
 
\(\displaystyle M\;=\;\frac{0.00956\;mol\;HCl}{0.05000\;L}\;=\;\mathbf{0.1911\;M\;HCl}\)

 

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Exercise 2. A 40.00 mL aqueous H₂SO₄ is titrated with KOH. It takes 18.35 mL of 0.325 M KOH to complete the reaction. What is the concentration of H₂SO₄?

First, write the balanced chemical equation.

H₂SO₄ (aq) + 2 KOH (aq) → K₂SO₄ (aq) + 2 H₂O (l)

There is a 1:2 mole ratio of acid to base.

\(\displaystyle mol\;KOH\;=\;0.01835\;L\times\frac{0.325\;M}{L}\;=\;0.00596\;moles\;KOH\)
 
Next find the number of moles of H₂SO₄ using the mole ratio from the balanced chemical equation.

\(\displaystyle 0.00596\;mol\;KOH\times\frac{1\;mol\;H_2SO_4}{2\;mol\;KOH}\;=\;0.00298\;mol\;H_2SO_4\)
 
Next we calculate the molarity of H₂SO₄.

\(\displaystyle \frac{0.00298\;mol\;H_2SO_4}{0.04000\;L}\;=\;\mathbf{0.0745\;M\;H_2SO_4}\)

 

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Exercise 3. Consider the following net ionic equation:

Cr₂O₇^2- (aq) + 6 Fe²⁺ (aq) + 14 H⁺ (aq) → 2 Cr³⁺ (aq) + 6 Fe³⁺ (aq) + 7 H₂O (l)

It takes 18.62 mL of 0.1625 M K₂Cr₂O₇ to tirate a sample of FeSO₄. What is the mass of the FeSO₄ sample?

First find the number of moles of Cr₂O₇^2-. Using mole ratios from the balanced chemical equation, calculate the moles of Fe²⁺. Once the moles of Fe²⁺ have been calculated, the grams of FeSO₄ can be determined. There are 0.1625 moles of Cr₂O₇^2- in 0.1625 M K₂Cr₂O₇.

\(\displaystyle 0.1625\;M\;Cr_2O_7^{2-}\times\;0.01862\;L\;=\;0.00303\;mol\;Cr2O_7^{2-}\)

Now, we use the mole ratio from the balanced equation to calculate the number of moles of Fe²⁺.

\(\displaystyle 0.00303\;mol\;Cr_2O_7^{2-}\;\times\frac{6\;mol\;Fe^{2+}}{1\;mol\;Cr_2O_7^{2-}}\;=\;0.0182\;mol\;Fe^{2+}\)
 
There is 1 mole of Fe²⁺ in 1 mole of FeSO₄.

\(\displaystyle 0.0182\;mol\;Fe^{2+}\;\times\;\frac{1\;mol\;FeSO_4}{1\;mol\;Fe^{2+}}\times\frac{151.908\;g\;FeSO_4}{1\;mol\;FeSO_4}\;=\;\mathbf{2.76\;g\;FeSO_4}\)

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Exercise 4. Consider the following equation:

I₂ (aq) + 2 S₂O₃^2- (aq) → S₄O₆^2- (aq) + 2 I^– (aq)

How many mL of 0.245 M Na₂S₂O₃ solution is needed to completely react with 2.252 g of I₂?

\(\displaystyle 2.252\;g\;I_2\;\times\frac{1\;mol}{253.8989\;g}\;=\;0.00887\;mol\;I_2\)

Using the mole ratio from the balanced chemical equation we calculate the number of moles of S₂O₃^2-.

\(\displaystyle 0.00887\;mol\;I_2\;\times\frac{2\;mol\;S_2O_3^{2-}}{1\;mol\;I_2}\;=\;0.0177\;mol\;S_2O_3^{2-}\)
 
There is one mole of S₂O₃^2- in 1 mole of Na₂S₂O₃

\(\displaystyle V\;=\;\frac{0.0177\;mol\;S_{2}O_{3}^{2-}}{0.245\;M}\;=\;0.0722\;L\;Na_{2}S_{2}O_{3}\;=\;\mathbf{72.2\;mL\;Na_{2}S_{2}O_{3}}\)

 

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Exercise 5. The amount of iron in an ore can be determined by oxidizing the sample with Br₂ to convert all of the iron to Fe³⁺. The Fe³⁺ is then titrated with Sn²⁺ to reduce the iron(III) ion to Fe³⁺ according to the following equation:

2 Fe³⁺ (aq) + Sn²⁺ (aq) → 2 Fe²⁺ (aq) + Sn⁴⁺ (aq)

Calculate the mass percent of Fe metal in a 0.1725 g sample of ore if 11.65 mL of 0.09505 M Sn²⁺ is used to titrate Fe³⁺.

\(\displaystyle mol\;Sn^{2+}\;=\;0.01165\;L\;\times\frac{0.09505\;mol}{L}\;=\;0.001107\;mol\;Sn^{2+}\)

\(\displaystyle 0.001107\;mol\;Sn^{2+}\;\times\frac{2\;mol\;Fe^{3+}}{1\;mol\;Sn^{2+}}\;=\;0.002214\;mol\;Fe^{3+}\)
 

The molar mass of Fe is 55.845 g/mol.

\(\displaystyle 0.002214\;mol\;Fe\;\times\;\frac{55.845\;g}{1\;mol}\;=\;0.1236\;g\;Fe\)
 
\(\displaystyle percent\;Fe\;=\;\frac{0.1236\;g\;Fe}{0.1725\;g\;ore}\;\times\;100\;=\;71.65\%\;Fe\)
 

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