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Chapter 4: Problem 104

The concentration of hydrogen peroxide in a solution is determined by titrating a \(10.0\) -mL sample of the solution with permanganate ion. $$ \begin{array}{r} 2 \mathrm{MnO}_{4}^{-}(a q)+5 \mathrm{H}_{2} \mathrm{O}_{2}(a q)+6 \mathrm{H}^{+}(a q) \longrightarrow \\ 2 \mathrm{Mn}^{2+}(a q)+5 \mathrm{O}_{2}(g)+8 \mathrm{H}_{2} \mathrm{O}(l) \end{array} $$ If it takes \(14.8 \mathrm{~mL}\) of \(0.134 \mathrm{M} \mathrm{MnO}_{4}^{-}\) solution to reach the equivalence point, what is the molarity of the hydrogen peroxide solution?

Short Answer

Expert verified
The concentration (molarity) of the hydrogen peroxide solution is \(0.4963\,M\).

Step by step solution

01

Identify the given data and what we need to find

We are given the information: - Volume of the hydrogen peroxide solution: \(10.0\,mL\) - Volume of the permanganate solution: \(14.8\,mL\) - Concentration of the permanganate solution: \(0.134\,M\) We need to find the molarity(concentration) of the hydrogen peroxide solution.
02

Convert volumes to liters

We need to convert the given volumes from milliliters to liters for further calculations using unit conversion factors: \(Volume_{H_2O_2} = 10.0\,mL \times \frac{1\,L}{1000\,mL} = 0.010\,L\) \(Volume_{MnO_4^-} = 14.8\,mL \times \frac{1\,L}{1000\,mL} = 0.0148\,L\)
03

Calculate the moles of permanganate ion

Now, we can find out the moles of permanganate ion (\(MnO_4^-\)) using the volume and concentration of the titrant solution: \(Moles_{MnO_4^-} = Volume_{MnO_4^-}\times Concentration_{MnO_4^-}\) \(Moles_{MnO_4^-} = 0.0148\,L \times 0.134\,M = 0.0019852\,mol\)
04

Use stoichiometry to find moles of hydrogen peroxide

Using the stoichiometry from the balanced chemical equation, we can find the moles of hydrogen peroxide: \(2\,MnO_4^-(aq) + 5\,H_2O_2(aq) + 6\,H^+(aq) \rightarrow 2\,Mn^{2+}(aq) + 5\,O_2(g) + 8\,H_2O(l)\) From the balanced equation, it is clear that \(2\,mol\,MnO_4^-\) are consumed for every \(5\,mol\,H_2O_2\): \(Moles_{H_2O_2} = \frac{5}{2} \times Moles_{MnO_4^-}\) \(Moles_{H_2O_2} = \frac{5}{2} \times 0.0019852\,mol = 0.004963\,mol\)
05

Calculate the concentration (molarity) of hydrogen peroxide

Now that we have the moles of hydrogen peroxide and the volume of the hydrogen peroxide solution, we can calculate the concentration (molarity) using the formula: \(Molarity_{H_2O_2} = \frac{Moles_{H_2O_2}}{Volume_{H_2O_2}}\) \(Molarity_{H_2O_2} = \frac{0.004963\,mol}{0.010\,L} = 0.4963\,M\) So, the concentration (M) of the hydrogen peroxide solution is \(0.4963\,M\).

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Most popular questions from this chapter

The safe drinking water standard for arsenic (which is usually found as arsenate, see \(4.115)\) is 50 parts per billion (ppb) in most developing countries. (a) How many grams of sodium arsenate are in 55 gallons of water, if the concentration of arsenate is 50 ppb? (b) In 1993 , naturally occurring arsenic was discovered as a major contaminant in the drinking water across the country of Bangladesh. Approximately 12 million people in Bangladesh still drink water from wells that have higher concentrations of arsenic than the standard. Recently, a chemistry professor from George Mason University was awarded a \(\$ 1\) million Grainger Challenge Prize for Sustainability for his development of a simple, inexpensive system for filtering naturally occuring arsenic from drinking water. The system uses buckets of sand, cast iron, activated carbon, and wood chips for trapping arsenic-containing minerals. Assuming the efficiency of such a bucket system is \(90 \%\) (meaning, \(90 \%\) of the arsenic that comes in is retained in the bucket and \(10 \%\) passes out of the bucket), how many times should water that is 500 ppb in arsenic be passed through to meet the 50 ppb standard?

Identify the precipitate (if any) that forms when the following solutions are mixed, and write a balanced equation for each reaction. (a) \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}\) and \(\mathrm{NaOH}\), (b) \(\mathrm{NaOH}\) and \(\mathrm{K}_{2} \mathrm{SO}_{4}\), (c) \(\mathrm{Na}_{2} \mathrm{~S}\) and \(\mathrm{Cu}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}\).

Which element is oxidized and which is reduced in the following reactions? (a) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) (b) \(3 \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2 \mathrm{Al}(s) \longrightarrow\) \(3 \mathrm{Fe}(s)+2 \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}(a q)\) (c) \(\mathrm{Cl}_{2}(a q)+2 \mathrm{NaI}(a q) \longrightarrow \mathrm{I}_{2}(a q)+2 \mathrm{NaCl}(a q)\) (d) \(\mathrm{PbS}(s)+4 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{PbSO}_{4}(s)+4 \mathrm{H}_{2} \mathrm{O}(l)\)

Complete and balance the following molecular equations, and then write the net ionic equation for each: (a) \(\mathrm{HBr}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(a q) \longrightarrow\) (b) \(\mathrm{Cu}(\mathrm{OH})_{2}(s)+\mathrm{HClO}_{4}(a q) \longrightarrow\) (c) \(\mathrm{Al}(\mathrm{OH})_{3}(\mathrm{~s})+\mathrm{HNO}_{3}(a q) \longrightarrow\)

Will precipitation occur when the following solutions are mixed? If so, write a balanced chemical equation for the reaction. (a) \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) and \(\mathrm{AgNO}_{3}\), (b) \(\mathrm{NaNO}_{3}\) and \(\mathrm{NiSO}_{4}\), (c) \(\mathrm{FeSO}_{4}\) and \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\).
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