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Chapter 5: Problem 50

The exact neutralization of \(10.00 \mathrm{mL}\) of \(0.1012 \mathrm{M}\) \(\mathrm{H}_{2} \mathrm{SO}_{4}(\text { aq })\) requires \(23.31 \mathrm{mL}\) of \(\mathrm{NaOH}\). What must be the molarity of the \(\mathrm{NaOH}(\mathrm{aq}) ?\) \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})+2 \mathrm{NaOH}(\mathrm{aq}) \longrightarrow\) \(\mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(1)\)

Short Answer

Expert verified
The molarity of the NaOH solution is 0.0868 M.

Step by step solution

01

Identify the Knowns

The following are identified from the problem: volume of H2SO4 solution = 10.00 mL = 0.01 L and molarity of H2SO4 = 0.1012 M; volume of NaOH = 23.31 mL = 0.02331 L. All volume values were converted from mL to L for consistency.
02

Calculate moles of H2SO4

Next, calculate the number of moles of H2SO4 reacted. The mole of a substance is its molarity times its volume in litres. Consider the relation: moles = molarity x volume. So, moles of H2SO4 = 0.1012 M x 0.01 L = 0.001012 mol.
03

Use stoichiometry to find moles of NaOH

In the balanced chemical equation, 2 moles of NaOH react with 1 mole of H2SO4. Therefore, 2 times the moles of H2SO4 will give the moles of NaOH. So, moles of NaOH = 2 x moles of H2SO4 = 2 x 0.001012 mol = 0.002024 mol.
04

Calculate molarity of NaOH

Molarity is defined as the number of moles per litre. The volume of NaOH solution used is 0.02331 L. Therefore, the molarity of NaOH is calculated as: Molarity of NaOH = moles of NaOH / volume of NaOH in L = 0.002024 mol / 0.02331 L = 0.0868 M

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Most popular questions from this chapter

Balance these equations for redox reactions occurring in basic solution. (a) \(\mathrm{CrO}_{4}^{2-}+\mathrm{S}_{2} \mathrm{O}_{4}^{2-} \longrightarrow \mathrm{Cr}(\mathrm{OH})_{3}(\mathrm{s})+\mathrm{SO}_{3}^{2-}\) (b) \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}+\mathrm{N}_{2} \mathrm{H}_{4} \longrightarrow\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}+\mathrm{N}_{2}(\mathrm{g})\) (c) \(\operatorname{Fe}(\mathrm{OH})_{2}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{s})\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{MnO}_{4}^{-} \longrightarrow\) \(\mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{MnO}_{2}(\mathrm{s})\)

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 solutions would you use to precipitate \(\mathrm{Mg}^{2+}\) from an aqueous solution of \(\mathrm{MgCl}_{2} ?\) Explain your choice. (a) \(\mathrm{KNO}_{3}(\mathrm{aq}) ;\) (b) \(\mathrm{NH}_{3}(\mathrm{aq}) ;\) (c) \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq});\) (d) \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(\mathrm{aq})\).

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?

An \(\mathrm{NaOH}(\mathrm{aq})\) solution cannot be made up to an exact concentration simply by weighing out the required mass of NaOH, because the NaOH is not pure. Also, water vapor condenses on the solid as it is being weighed. The solution must be standardized by titration. For this purpose, a 25.00 \(\mathrm{mL}\) sample of an NaOH(aq) solution requires 28.34 \(\mathrm{mL}\) of 0.1085 \(\mathrm{M}\) HCl. What is the molarity of the NaOH(aq)? \(\mathrm{HCl}(\mathrm{aq})+\mathrm{NaOH}(\mathrm{aq}) \longrightarrow \mathrm{NaCl}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(1)\)
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