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Chapter 4: Problem 88

Would the volume of a \(0.10 M \mathrm{NaOH}\) solution needed to titrate \(25.0 \mathrm{~mL}\) of a \(0.10 \mathrm{M} \mathrm{HNO}_{2}\) (a weak acid) solution be different from that needed to titrate \(25.0 \mathrm{~mL}\) of a \(0.10 \mathrm{M} \mathrm{HCl}\) (a strong acid) solution?

Short Answer

Expert verified
No, the volume of a 0.10 M \(NaOH\) solution needed to titrate 25.0 mL of a 0.10 M \(HNO_2\) solution would not be different from that needed to titrate 25.0 mL of a 0.10 M \(HCl\) solution. In both cases, the volume required is 25.0 mL.

Step by step solution

01

Understand the Neutralization Reaction

In a neutralization reaction between an acid and base, the resulting products are water and a salt. Here the base is \(NaOH\) and the acids are \(HNO_2\) and \(HCl\). The reactions are: \(HNO_2 + NaOH \rightarrow NaNO_2 + H_2O\) and \(NaOH+HCl \rightarrow NaCl + H_2O\). It is seen that the stoichiometry of both reactions is a 1 to 1 molar ratio.
02

Calculate the Moles of Acid

For both the acids \(HNO_2\) and \(HCl\), the moles can be calculated using the equation n = Molarity x Volume. Here, molarity is 0.10M and volume is 25.0mL (converted to L which is 0.025L). Thus, the moles of acid = 0.10M x 0.025L = 0.0025 mol.
03

Determine the Moles of Base

Because the molar ratio of acid to base in the neutralization reaction is 1:1, the moles of \(NaOH\) required for neutralization will also be 0.0025 mol.
04

Determine the Volume of Base

To find the volume of \(NaOH\) solution needed for the titration, use the equation Volume = moles/Molarity. In this case, the volume of \(NaOH\) = 0.0025 mol/ 0.10 M = 0.025L or 25.0mL. This volume is required for titrating both the \(HNO_2\) and \(HCl\) solutions.

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Most popular questions from this chapter

Chlorine forms a number of oxides with the following oxidation numbers: \(+1,+3,+4,+6,\) and +7 . Write a formula for each of these compounds.

A quantity of \(25.0 \mathrm{~mL}\) of a solution containing both \(\mathrm{Fe}^{2+}\) and \(\mathrm{Fe}^{3+}\) ions is titrated with \(23.0 \mathrm{~mL}\) of \(0.0200 M \mathrm{KMnO}_{4}\) (in dilute sulfuric acid). As a result, all of the \(\mathrm{Fe}^{2+}\) ions are oxidized to \(\mathrm{Fe}^{3+}\) ions. Next, the solution is treated with Zn metal to convert all of the \(\mathrm{Fe}^{3+}\) ions to \(\mathrm{Fe}^{2+}\) ions. Finally, the solution containing only the \(\mathrm{Fe}^{2+}\) ions requires \(40.0 \mathrm{~mL}\) of the same \(\mathrm{KMnO}_{4}\) solution for oxidation to \(\mathrm{Fe}^{3+}\). Calculate the molar concentrations of \(\mathrm{Fe}^{2+}\) and \(\mathrm{Fe}^{3+}\) in the original solution. The net ionic equation is \(\mathrm{MnO}_{4}^{-}+5 \mathrm{Fe}^{2+}+8 \mathrm{H}^{+} \longrightarrow\) \(\mathrm{Mn}^{2+}+5 \mathrm{Fe}^{3+}+4 \mathrm{H}_{2} \mathrm{O}\)

A useful application of oxalic acid is the removal of rust \(\left(\mathrm{Fe}_{2} \mathrm{O}_{3}\right)\) from, say, bathtub rings according to the reaction \(\begin{aligned} \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+& 6 \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(a q) \longrightarrow \\ & 2 \mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}^{3-}(a q)+3 \mathrm{H}_{2} \mathrm{O}+6 \mathrm{H}^{+}(a q) \end{aligned}\) Calculate the number of grams of rust that can be removed by \(5.00 \times 10^{2} \mathrm{~mL}\) of a \(0.100 \mathrm{M}\) solution of oxalic acid.

On the basis of oxidation number considerations, one of the following oxides would not react with molecular oxygen: \(\mathrm{NO}, \mathrm{N}_{2} \mathrm{O}, \mathrm{SO}_{2}, \mathrm{SO}_{3}, \mathrm{P}_{4} \mathrm{O}_{6}\) Which one is it? Why?

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