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Chapter 5: Problem 60

The electrolyte in a lead storage battery must have a concentration between 4.8 and \(5.3 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) if the battery is to be most effective. A \(5.00 \mathrm{mL}\) sample of a battery acid requires \(49.74 \mathrm{mL}\) of \(0.935 \mathrm{M} \mathrm{NaOH}\) for its complete reaction (neutralization). Does the concentration of the battery acid fall within the desired range? [Hint: Keep in mind that the \(\mathrm{H}_{2} \mathrm{SO}_{4}\) produces two \(\mathrm{H}^{+}\) ions per formula unit.]

Short Answer

Expert verified
The concentration of the battery acid (4.66 M) does not fall within the desired range of 4.8-5.3 M for the battery to be most effective.

Step by step solution

01

Determine moles of Sodium Hydroxide

Based on the volume and molarity of sodium hydroxide in the reaction, calculate the number of moles by using the formula: Molarity = Moles / Volume. Rearranging this formula to find moles gives: Moles = Molarity * Volume. Remember to convert volume from mL to L. So, moles of \(\mathrm{NaOH} = 0.935 \mathrm{M}*\frac{49.74 \mathrm{mL}}{1000 \mathrm{mL/L}}=0.0465 \mathrm{mol}\)
02

Determine moles of Sulfuric Acid

Knowing that every mole of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) reacts with two moles of \(\mathrm{NaOH}\), we can calculate the moles of sulfuric acid. So, moles of \(\mathrm{H}_{2} \mathrm{SO}_{4} = \frac{0.0465 \mathrm{mol}}{2} = 0.0233 \mathrm{mol}\)
03

Calculate concentration of Sulfuric Acid

The molarity (concentration in mol/L) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is calculated using the formula: Molarity = Moles / Volume. Volume of acid sample is 5.00 mL, so convert this to litres: Volume = \(\frac{5.00 \mathrm{mL}}{1000 \mathrm{mL/L}}\). Hence, the molarity of sulfuric acid: \(\mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4} = \frac{0.0233 \mathrm{mol}}{0.005 \mathrm{L}} = 4.66 \mathrm{M}\)
04

Compare Concentration with Desired Range

The molarity found is now compared to the effective range given in the problem (4.8 - 5.3M). Since 4.66 lies outside this range, it can be concluded that the concentration of the battery acid does not fall within the desired range.

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Most popular questions from this chapter

A \(7.55 \mathrm{g}\) sample of \(\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s})\) is added to \(125 \mathrm{mL}\) of a vinegar that is \(0.762 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH} .\) Will the resulting solution still be acidic? Explain.

Iron (Fe) is obtained from rock that is extracted from open pit mines and then crushed. The process used to obtain the pure metal from the crushed rock produces solid waste, called tailings, which are stored in disposal areas near the mines. The tailings pose a serious environmental risk because they contain sulfides, such as pyrite ( \(\mathrm{FeS}_{2}\) ), which oxidize in air to produce metal ions and \(\mathrm{H}^{+}\) ions that can enter into surface water or ground water. The oxidation of \(\mathrm{FeS}_{2}\) to \(\mathrm{Fe}^{3+}\) is described by the unbalanced chemical equation below. \(\mathrm{FeS}_{2}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow\) \(\quad \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq}) \quad(\text { not balanced })\) Thus, the oxidation of pyrite produces \(\mathrm{Fe}^{3+}\) and \(\mathrm{H}^{+}\) ions that can leach into surface or ground water. The leaching of \(\mathrm{H}^{+}\) ions causes the water to become very acidic. To prevent acidification of nearby ground or surface water, limestone \(\left(\mathrm{CaCO}_{3}\right)\) is added to the tailings to neutralize the \(\mathrm{H}^{+}\) ions: \(\mathrm{CaCO}_{3}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \underset{\mathrm{Ca}^{2+}}{\longrightarrow}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{CO}_{2}(\mathrm{g})\) (a) Balance the equation above for the reaction of \(\mathrm{FeS}_{2}\) and \(\mathrm{O}_{2}\). [ Hint: Start with the half-equations \(\mathrm{FeS}_{2}(\mathrm{s}) \rightarrow\) \(\left.\mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq}) \text { and } \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{H}_{2} \mathrm{O}(1) .\right]\) (b) What is the minimum amount of \(\mathrm{CaCO}_{3}(\mathrm{s})\) required, per kilogram of tailings, to prevent contamination if the tailings contain \(3 \%\) S by mass? Assume that all the sulfur in the tailings is in the form \(\mathrm{FeS}_{2}\).

A method of producing phosphine, \(\mathrm{PH}_{3}\), from elemental phosphorus, \(P_{4}\), involves heating the \(P_{4}\) with \(\mathrm{H}_{2} \mathrm{O} .\) An additional product is phosphoric acid, \(\mathrm{H}_{3} \mathrm{PO}_{4}\) Write a balanced equation for this reaction.

What is the net ionic equation for the reaction that occurs when an aqueous solution of \(\mathrm{KI}\) is added to an aqueous solution of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2} ?\)

\(\mathrm{NH}_{3}(\mathrm{aq})\) conducts electric current only weakly. The same is true for \(\mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq}) .\) When these solutions are mixed, however, the resulting solution is a good conductor. How do you explain this?
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