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Chapter 4: Problem 79

A 25.00-mL sample of hydrochloric acid solution requires 24.16 mL of 0.106 M sodium hydroxide for complete neutralization. What is the concentration of the original hydrochloric acid solution?

Short Answer

Expert verified
The concentration of the original hydrochloric acid solution is \(0.1026 M\).

Step by step solution

01

Write the balanced chemical equation for the reaction

For the neutralization reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH), the balanced chemical equation is: \[HCl(aq) + NaOH(aq) \rightarrow NaCl(aq) + H_2O(l)\] The stoichiometric ratio between HCl and NaOH is 1:1.
02

Determine the moles of NaOH

We know the volume and concentration of the NaOH solution, which can be used to calculate the moles of NaOH present in the solution. Use the following formula to find moles of NaOH: Moles = Molarity × Volume The molarity of NaOH is given as 0.106 M, and the volume of NaOH is 24.16 mL. Note that to do the calculation, you must convert mL to L by dividing by 1000. Moles of NaOH = (0.106 moles/L) × (24.16 mL × (1 L / 1000 mL)) Moles of NaOH = 0.00256656 moles
03

Determine the moles of HCl

According to the stoichiometric ratio, the moles of HCl and NaOH have a 1:1 ratio. Therefore, the moles of HCl is equal to the moles of NaOH. Moles of HCl = 0.00256656 moles
04

Calculate the concentration of HCl

We know the moles of HCl and the volume of the HCl solution (25.00 mL). We can now calculate the concentration (molarity) of HCl using the formula: Molarity = Moles / Volume Convert the volume of HCl from mL to L by dividing by 1000. Volume of HCl in L = 25.00 mL × (1 L / 1000 mL) = 0.025 L Now calculate the molarity of HCl: Molarity of HCl = (0.00256656 moles) / (0.025 L) Molarity of HCl = 0.1026 M Therefore, the concentration of the original hydrochloric acid solution is 0.1026 M.

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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 230 -mL sample of a \(0.275-M \mathrm{CaCl}_{2}\) solution is left on a hot plate overnight; the following morning, the solution is 1.10\(M .\) What volume of water evaporated from the 0.275\(M \mathrm{CaCl}_{2}\) solution?

Balance the following oxidation–reduction reactions that occur in acidic solution using the half-reaction method. a. $\mathrm{Cu}(s)+\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{NO}(g)$ b. $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{Cl}^{-}(a q) \rightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{Cl}_{2}(g)$ c. $\mathrm{Pb}(s)+\mathrm{PbO}_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{PbSO}_{4}(s)$ d. $\mathrm{Mn}^{2+}(a q)+\mathrm{NaBiO}_{3}(s) \rightarrow \mathrm{Bi}^{3+}(a q)+\mathrm{MnO}_{4}-(a q)$ e. $\mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{Zn}(s) \rightarrow \mathrm{AsH}_{3}(g)+\mathrm{Zn}^{2+}(a q)$

A mixture contains only NaCl and \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}\) . A \(0.456-\mathrm{g}\) sample of the mixture is dissolved in water, and an excess of NaOH is added, producing a precipitate of \(\mathrm{Fe}(\mathrm{OH})_{3}\) . The precipitate is filtered, dried, and weighed. Its mass is 0.107 \(\mathrm{g}\) . Calculate the following. a. the mass of iron in the sample b. the mass of Fe(NO \(_{3} )_{3}\) in the sample c. the mass percent of \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}\) in the sample

When organic compounds containing sulfur are bumed, sulfur dioxide is produced. The amount of \(\mathrm{SO}_{2}\) formed can be determined by the reaction with hydrogen peroxide: $$\mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{SO}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{4}(a q)$$ The resulting sulfuric acid is then titrated with a standard NaOH solution. A 1.302 -g sample of coal is burned and the \(\mathrm{SO}_{2}\) is collected in a solution of hydrogen peroxide. It took 28.44 \(\mathrm{mL}\) of a $0.1000-M \mathrm{NaOH}$ solution to titrate the resulting sulfuric acid. Calculate the mass percent of sulfur in the coal sample. Sulfuric acid has two acidic hydrogens.
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