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Chapter 5: Problem 88

Chile saltpeter is a natural source of \(\mathrm{NaNO}_{3}\); it also contains \(\mathrm{NaIO}_{3} .\) The \(\mathrm{NaIO}_{3}\) can be used as a source of iodine. Iodine is produced from sodium iodate in a two-step process occurring under acidic conditions: \(\begin{aligned} \mathrm{IO}_{3}^{-}(\mathrm{aq})+\mathrm{HSO}_{3}^{-}(\mathrm{aq}) & \longrightarrow \mathrm{I}^{-}(\mathrm{aq}) +\mathrm{SO}_{4}^{2-}(\mathrm{aq}) \end{aligned} \quad\) ( not balanced) \(\mathrm{I}^{-}(\mathrm{aq})+\mathrm{IO}_{3}^{-}(\mathrm{aq}) \longrightarrow\) \(\mathrm{I}_{2}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad(\text { not balanced })\) In the illustration, a 5.00 L sample of a \(\mathrm{NaIO}_{3}(\mathrm{aq})\) solution containing \(5.80 \mathrm{g} \mathrm{NaIO}_{3} / \mathrm{L}\) is treated with the stoichiometric quantity of \(\mathrm{NaHSO}_{3}\) (no excess of either reactant). Then, a further quantity of the initial \(\mathrm{NaIO}_{3}(\mathrm{aq})\) is added to the reaction mixture to bring about the second reaction. (a) How many grams of NaHSO \(_{3}\) are required in the first step? (b) What additional volume of the starting solution must be added in the second step?

Short Answer

Expert verified
For the first reaction, 15.3 g of NaHSO3 are needed. An additional 1.00 L of the initial NaIO3 solution is needed for the second reaction.

Step by step solution

01

Balance the given equations

The balanced equations are:\n\(\mathrm{IO}_{3}^{-}(\mathrm{aq})+\mathrm{HSO}_{3}^{-}(\mathrm{aq}) \longrightarrow \mathrm{I}^{-}(\mathrm{aq}) +\mathrm{SO}_{4}^{2-}(\mathrm{aq}+2\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) and \(\mathrm{5I}^{-}(\mathrm{aq})+\mathrm{IO}_{3}^{-}(\mathrm{aq}) + 6 H^{+} \longrightarrow 3 \mathrm{I}_{2}(\mathrm{s})+3\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\).\n
02

Calculate molar mass of NaIO3 and NaHSO3

The molecular mass of NaIO3 is \(85.0+15.999 \times 3+15.999 \times 3=197.9 \, g/mol\).\nThe molecular mass of NaHSO3 is \(22.99+1.008+32.06+3 \times 15.999 = 104.06 \, g/mol\).
03

Find the moles of NaIO3

The number of moles of NaIO3 in the 5.00 L solution can be calculated as \(5.00 L \times \frac{5.80g}{L} \times \frac{1 mol}{197.9 g} = 0.147 mol\).
04

Calculate the mass of NaHSO3 needed

From stoichiometry of reaction (1), for every 1 mole of IO3- we need 1 mole of HSO3-. Therefore, we need 0.147 moles of NaHSO3. This corresponds to a mass of \(0.147 mol \times 104.06 g/mol = 15.3 g\).
05

Calculate Additional volume of starting solution needed

For the second reaction, we need 1 mole of IO3- for every 5 moles of I-. In the first step, we produced 0.147 moles of I-, therefore, we need \(0.147 mol/5 = 0.0294 mol\) of IO3-. The volume of the starting solution containing this quantity is given by \(0.0294 mol \times \frac{197.9 g}{mol} \times \frac{L}{5.80 g} = 1.00 L\).

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Most popular questions from this chapter

What volume of \(0.0962 \mathrm{M} \mathrm{NaOH}\) is required to exactly neutralize \(10.00 \mathrm{mL}\) of \(0.128 \mathrm{M} \mathrm{HCl} ?\)

Phosphorus is essential for plant growth, but an excess of phosphorus can be catastrophic in aqueous ecosystems. Too much phosphorus can cause algae to grow at an explosive rate and this robs the rest of the ecosystem of oxygen. Effluent from sewage treatment plants must be treated before it can be released into lakes or streams because the effluent contains significant amounts of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\). (Detergents are a major contributor to phosphorus levels in domestic sewage because many detergents contain \(\mathrm{Na}_{2} \mathrm{HPO}_{4}\) ) A simple way to remove \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\) from the effluent is to treat it with lime, \(\mathrm{CaO}\) which produces \(\mathrm{Ca}^{2+}\) and \(\mathrm{OH}^{-}\) ions in water. The \(\mathrm{OH}^{-}\) ions convert \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\) ions into \(\mathrm{PO}_{4}^{3-}\) ions and, finally, \(\mathrm{Ca}^{2+}, \mathrm{OH}^{-}\), and \(\mathrm{PO}_{4}^{3-}\) ions combine to form a precipitate of \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}(\mathrm{s})\) (a) Write balanced chemical equations for the four reactions described above. [Hint: The reactants are \(\mathrm{CaO}\) and \(\mathrm{H}_{2} \mathrm{O} ; \mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\left.\mathrm{OH}^{-} ; \mathrm{HPO}_{4}^{2-} \text { and } \mathrm{OH}^{-} ; \mathrm{Ca}^{2+}, \mathrm{PO}_{4}^{3-}, \text { and } \mathrm{OH}^{-} .\right]\) (b) How many kilograms of lime are required to remove the phosphorus from a \(1.00 \times 10^{4}\) L holding tank filled with contaminated water, if the water contains \(10.0 \mathrm{mg}\) of phosphorus per liter?

For use in titrations, we want to prepare \(20 \mathrm{L}\) of \(\mathrm{HCl}(\mathrm{aq})\) with a concentration known to four significant figures. This is a two-step procedure beginning with the preparation of a solution of about \(0.10 \mathrm{M}\) HCl. A sample of this dilute HCl(aq) is titrated with a NaOH(aq) solution of known concentration. (a) How many milliliters of concentrated \(\mathrm{HCl}(\mathrm{aq})\) \((d=1.19 \mathrm{g} / \mathrm{mL} ; 38 \% \mathrm{HCl}, \text { by mass })\) must be diluted with water to 20.0 L to prepare \(0.10 \mathrm{M} \mathrm{HCl}\) ? (b) \(\mathrm{A} 25.00\) \(\mathrm{mL}\) sample of the approximately \(0.10\) \(\mathrm{M}\) HCl prepared in part (a) requires \(20.93\) \(\mathrm{mL}\) of \(0.1186\) \(\mathrm{M}\) NaOH for its titration. What is the molarity of the \(\mathrm{HCl}(\mathrm{aq}) ?\) (c) Why is a titration necessary? That is, why not prepare a standard solution of \(0.1000\) \(\mathrm{M} \mathrm{HCl}\) simply by an appropriate dilution of the concentrated HCl(aq)?

Which aqueous solution has the greatest \(\left[\mathrm{H}^{+}\right]:\) (a) \(0.011 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH} ;\) (b) \(0.010 \mathrm{M} \mathrm{HCl} ;\) (c) \(0.010 \mathrm{M}\) \(\mathrm{H}_{2} \mathrm{SO}_{4} ;\) (d) \(1.00 \mathrm{M} \mathrm{NH}_{3} ?\) Explain your choice.

We want to determine the acetylsalicyclic acid content of a series of aspirin tablets by titration with \(\mathrm{NaOH}(\mathrm{aq})\) Each of the tablets is expected to contain about \(0.32\) \(\mathrm{g}\) of \(\mathrm{HC}_{9} \mathrm{H}_{7} \mathrm{O}_{4} \cdot\) What molarity of \(\mathrm{NaOH}(\mathrm{aq})\) should we use for titration volumes of about \(23\) \(\mathrm{mL}\) ? (This procedure ensures good precision and allows the titration of two samples with the contents of a 50 mL buret.) \(\mathrm{HC}_{9} \mathrm{H}_{7} \mathrm{O}_{4}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \longrightarrow_{\mathrm{C}_{9} \mathrm{H}_{7} \mathrm{O}_{4}^{-}}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(1)\)
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