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Chapter 4: Problem 141

You are given a solid that is a mixture of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) and \(\mathrm{K}_{2} \mathrm{SO}_{4}\) A 0.205-g sample of the mixture is dissolved in water. An excess of an aqueous solution of \(\mathrm{BaCl}_{2}\) is added. The BaSO\(_{4}\) that is formed is filtered, dried, and weighed. Its mass is 0.298 g. What mass of \(\mathrm{SO}_{4}^{2-}\) ion is in the sample? What is the mass percent of \(\mathrm{SO}_{4}^{2-}\) ion in the sample? What are the percent compositions by mass of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) and \(\mathrm{K}_{2} \mathrm{SO}_{4}\) in the sample?

Short Answer

Expert verified
The mass of \(\mathrm{SO}_{4}^{2-}\) ion in the 0.205-g sample is 0.1227 g. The mass percent of \(\mathrm{SO}_{4}^{2-}\) ion in the sample is 59.85%. The combined percent compositions by mass of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) and \(\mathrm{K}_{2} \mathrm{SO}_{4}\) in the sample is 40.15%.

Step by step solution

01

Find moles of BaSO\(_{4}\) formed

To find the moles of BaSO\(_{4}\) formed, we use the mass of BaSO\(_{4}\) and its molar mass. The molar mass of BaSO\(_{4}\) is calculated as follows: \[M_{\text{BaSO}_4} = M_{\text{Ba}} + M_{\text{S}} + 4 \cdot M_{\text{O}}\] Using the atomic masses from the periodic table: \[M_{\text{BaSO}_4} = 137.33 \,(\text{Ba}) + 32.07 \,(\text{S}) + 4 \cdot 16.00 \,(\text{O}) = 233.43\, \text{g/mol}\] Now we can find the moles of BaSO\(_{4}\) formed: \[n_{\text{BaSO\(_4\)}}= \frac{m_{\text{BaSO\(_4\)}}}{M_{\text{BaSO}_4}} = \frac{0.298 \, \text{g}}{233.43\, \text{g/mol}} = 1.277 \times 10^{-3}\, \text{mol}\]
02

Find moles of \(\mathrm{SO}_{4}^{2-}\) ion in the sample

Since each mole of BaSO\(_{4}\) contains one mole of \(\mathrm{SO}_{4}^{2-}\) ion, the moles of \(\mathrm{SO}_{4}^{2-}\) ion in the sample are equal to the moles of BaSO\(_{4}\) formed. Therefore, the moles of \(\mathrm{SO}_{4}^{2-}\) ion are: \[n_{\text{SO\(_4^{2-}\)}} = n_{\text{BaSO\(_4\)}} = 1.277 \times 10^{-3}\, \text{mol}\]
03

Find the mass of \(\mathrm{SO}_{4}^{2-}\) ion in the sample

To find the mass of \(\mathrm{SO}_{4}^{2-}\) ion in the sample, we use its molar mass and the moles we found in step 2. The molar mass of \(\mathrm{SO}_{4}^{2-}\) ion is: \[M_{\text{SO\(_4^{2-}\)}} = M_{\text{S}} + 4 \cdot M_{\text{O}} = 32.07 \,(\text{S}) + 4 \cdot 16.00 \,(\text{O}) = 96.07\, \text{g/mol}\] Now, we can find the mass of \(\mathrm{SO}_{4}^{2-}\) ion in the sample as follows: \[m_{\text{SO\(_4^{2-}\)}} = n_{\text{SO\(_4^{2-}\)}} \cdot M_{\text{SO\(_4^{2-}\)}} = (1.277 \times 10^{-3}\, \text{mol}) (96.07\, \text{g/mol}) = 0.1227\, \text{g}\]
04

Calculate mass percent of \(\mathrm{SO}_{4}^{2-}\) ion in the sample

To find the mass percent of \(\mathrm{SO}_{4}^{2-}\) ion in the sample, we use the mass of \(\mathrm{SO}_{4}^{2-}\) ion found in step 3 and the given mass of the sample (0.205 g), then apply the mass percent formula: \[\text{Mass percent of SO\(_4^{2-}\)} = \frac{m_{\text{SO\(_4^{2-}\)}}}{m_{\text{sample}}} \cdot 100\%\] \[= \frac{0.1227\, \text{g}}{0.205\, \text{g}} \cdot 100\% = 59.85\%\]
05

Calculate percent compositions by mass of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) and \(\mathrm{K}_{2} \mathrm{SO}_{4}\)

Since the mass percent of \(\mathrm{SO}_{4}^{2-}\) ion in the sample is 59.85%, the remaining mass percent (i.e., 100% - 59.85% = 40.15%) must be contributed by the \(\mathrm{Na}^+\) and \(\mathrm{K}^+\) ions combined. The percent compositions by mass of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) and \(\mathrm{K}_{2} \mathrm{SO}_{4}\) are not given separately. We can find their combined mass percent by adding their mass percents, which is equal to the mass percent of \(\mathrm{SO}_{4}^{2-}\) we found in step 4. Therefore, the combined percent compositions by mass of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) and \(\mathrm{K}_{2} \mathrm{SO}_{4}\) in the sample is 40.15%.

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Most popular questions from this chapter

Chromium has been investigated as a coating for steel cans. The thickness of the chromium film is determined by dissolving a sample of a can in acid and oxidizing the resulting \(\mathrm{Cr}^{3+}\) to $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$ with the peroxydisulfate ion: $$\begin{aligned} \mathrm{S}_{2} \mathrm{O}_{\mathrm{x}}^{2-}(a q)+\mathrm{Cr}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \\ &+\mathrm{SO}_{4}^{2-}(a q)+\mathrm{H}^{+}(a q)(\text { Unbalanced }) \end{aligned} $$ After removal of unreacted \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-},\) an excess of ferrous ammonium sulfate $\left[\mathrm{Fe}\left(\mathrm{NH}_{4}\right)_{2}\left(\mathrm{SO}_{4}\right)_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\right]$ is added, reacting with \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) produced from the first reaction. The unreacted \(\mathrm{Fe}^{2+}\) from the excess ferrous ammonium sulfate is titrated with a separate \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) solution. The reaction is: $$\begin{array}{c}{\mathrm{H}^{+}(a q)+\mathrm{Fe}^{2+}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \longrightarrow \mathrm{Fe}^{3+}(a q)} \\\ {+\mathrm{Cr}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(I) \text { (Unbalanced) }}\end{array}$$ a. Write balanced chemical equations for the two reactions. b. In one analysis, a \(40.0-\mathrm{cm}^{2}\) sample of a chromium-plated can was treated according to this procedure. After dissolution and removal of excess \(\mathrm{S}_{2} \mathrm{O}_{8}^{2}-, 3.000 \mathrm{g}\) of $\operatorname{Fe}\left(\mathrm{NH}_{4}\right)_{2}\left(\mathrm{SO}_{4}\right)_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\( was added. It took 8.58 \)\mathrm{mL}$ of 0.0520 \(\mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) solution to completely react with the excess \(\mathrm{Fe}^{2+}\) . Calculate the thickness of the chromium film on the can. (The density of chromium is 7.19 $\mathrm{g} / \mathrm{cm}^{3} .$ )

A student titrates an unknown amount of potassium hydrogen phthalate $\left(\mathrm{KHC}_{8} \mathrm{H}_{4} \mathrm{O}_{4}, \text { often abbreviated } \mathrm{KHP}\right)\( with 20.46 \)\mathrm{mL}$ of a 0.1000-M NaOH solution. KHP (molar mass 5 204.22 g/ mol) has one acidic hydrogen. What mass of KHP was titrated (reacted completely) by the sodium hydroxide solution?

Balance the following oxidation–reduction reactions that occur in acidic solution using the half-reaction method. a. $\mathbf{I}^{-}(a q)+\mathrm{ClO}^{-}(a q) \rightarrow \mathrm{I}_{3}^{-}(a q)+\mathrm{Cl}^{-}(a q)$ b. $\mathrm{As}_{2} \mathrm{O}_{3}(s)+\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{NO}(g)$ c. $\mathrm{Br}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{Br}_{2}(l)+\mathrm{Mn}^{2+}(a q)$ d. $\mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \rightarrow \mathrm{CH}_{2} \mathrm{O}(a q)+\mathrm{Cr}^{3+}(a q)$

Assign the oxidation state for the element listed in each of the following compounds: \(\mathrm{S}\) in \(\mathrm{MgSO}_{4}\)_______ \(\mathrm{Pb}\) in \(\mathrm{PbSO}_{4}\)______ \(\mathrm{O}\) in \(\mathrm{O}_{2}\)___________ \(\mathrm{Ag}\) in Ag _________________________ \(\mathrm{Cu}\) in \(\mathrm{CuCl}_{2}\)_______

You are given a 1.50-g mixture of sodium nitrate and sodium chloride. You dissolve this mixture into 100 mL of water and then add an excess of 0.500 M silver nitrate solution. You produce a white solid, which you then collect, dry, and measure. The white solid has a mass of 0.641 g. a. If you had an extremely magnified view of the solution (to the atomic- molecular level), list the species you would see (include charges, if any). b. Write the balanced net ionic equation for the reaction that produces the solid. Include phases and charges. c. Calculate the mass percent of sodium chloride in the original unknown mixture.
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