Stoichiometry and Redox Titration
Once a redox equation is balanced, its coefficients mean exactly what they mean in any other reaction: mole ratios. What is different in practice is how these reactions are often used. Chemists run them drop by drop from a buret until a color change shows that the substance being analyzed is completely used up.
Moles from a Balanced Redox Equation
Imagine a lab where you add purple $\mathrm{KMnO_4}$ solution drop by drop to a flask of $\mathrm{Fe^{2+}}$ in acid. The color disappears instantly, until one single drop leaves a faint pink that lasts. How many moles of iron(II) reacted with the volume you read on the buret?
Before using a ratio, you need a balanced equation. In acid, balance with half-reactions: balance each half-reaction for atoms (add $\mathrm{H_2O}$, then $\mathrm{H^+}$) and charge (add $\mathrm{e^-}$), multiply so the electrons match, then add them.
In the permanganate titration of dissolved iron, the permanganate ion quantitatively oxidizes the iron(II) cation in a strongly acidic medium produced with excess…
In the permanganate titration of dissolved iron, the permanganate ion quantitatively oxidizes the iron(II) cation in a strongly acidic medium produced with excess sulfuric acid, being reduced to the manganese(II) ion while the iron is oxidized to the iron(III) ion.
- a) Balance, using the half-reaction (ion-electron) method, the overall ionic equation for this redox process.
- b) If exactly $12.50\,\mathrm{mL}$ of an aqueous potassium permanganate solution $0.0200\,\mathrm{mol\times L^{-1}}$ is used to titrate a test sample, determine the amount of substance (in moles) of $\mathrm{Fe^{2+}}$ ions present in the sample analyzed.
Permanganate Titrations
In a buret, you read the liquid level before and after, and the titrant volume is the difference. For example, readings of $3.10\ \text{mL}$ and $18.10\ \text{mL}$ mean $15.00\ \text{mL}$ were delivered.
The titration of oxalic acid with potassium permanganate requires controlled prior heating of the Erlenmeyer flask to about $60\,^\circ\mathrm{C}$ to speed up the slow…
The titration of oxalic acid with potassium permanganate requires controlled prior heating of the Erlenmeyer flask to about $60\,^\circ\mathrm{C}$ to speed up the slow initial kinetics before the $\mathrm{Mn^{2+}}$ ion produced acts as an autocatalyst. The balanced reaction is:
$2\,\mathrm{MnO_4^-} + 5\,\mathrm{H_2C_2O_4} + 6\,\mathrm{H^+} \rightarrow 2\,\mathrm{Mn^{2+}} + 10\,\mathrm{CO_2} + 8\,\mathrm{H_2O}$
$20.00\,\mathrm{mL}$ of a standard solution of $\mathrm{H_2C_2O_4}$ $0.0500\,\mathrm{mol\times L^{-1}}$ is titrated, using $16.00\,\mathrm{mL}$ of $\mathrm{KMnO_4}$ solution. Calculate the molar concentration of the potassium permanganate solution used.
You have an aliquot of $25.0\,\mathrm{mL}$ of an aqueous solution containing $\mathrm{Fe^{2+}}$, treated with excess sulfuric acid. It is titrated by permanganate…
You have an aliquot of $25.0\,\mathrm{mL}$ of an aqueous solution containing $\mathrm{Fe^{2+}}$, treated with excess sulfuric acid. It is titrated by permanganate titration with a standard solution of $\mathrm{KMnO_4}$ $0.0200\,\mathrm{mol\times L^{-1}}$. In the experimental diagram below you can see the initial and final meniscus readings of the solution in the buret, corresponding to $3.10\,\mathrm{mL}$ and $18.10\,\mathrm{mL}$ respectively:
Consider the balanced ionic reaction:
$\mathrm{MnO_4^-} + 5\,\mathrm{Fe^{2+}} + 8\,\mathrm{H^+} \rightarrow \mathrm{Mn^{2+}} + 5\,\mathrm{Fe^{3+}} + 4\,\mathrm{H_2O}$
- a) Determine the molar concentration of $\mathrm{Fe^{2+}}$ ions in the solution analyzed.
- b) Justify the experimental criterion that defines the end point in this permanganate titration and state why hydrochloric acid should not be used instead of sulfuric acid to acidify the medium.
Iodometry and Indirect Titration
When you mix colorless bleach with a colorless potassium iodide solution, the liquid instantly turns dark brown because iodine forms. How can that brown product help you measure the hypochlorite, if you never titrate the hypochlorite directly?
In quantitative chemical analysis, a "blank test" is often carried out to correct for the volume of reagent consumed by impurities in the solvents or by the matrix…
In quantitative chemical analysis, a "blank test" is often carried out to correct for the volume of reagent consumed by impurities in the solvents or by the matrix itself.
In an iodometric titration with sodium thiosulfate according to the reaction:
$\mathrm{I_2} + 2\,\mathrm{S_2O_3^{2-}} \rightarrow 2\,\mathrm{I^-} + \mathrm{S_4O_6^{2-}}$
a test solution is titrated using $24.35\,\mathrm{mL}$ of $\mathrm{Na_2S_2O_3}$ $0.0100\,\mathrm{mol\times L^{-1}}$. A blank test with all the reagents except the analyte requires $0.15\,\mathrm{mL}$ of the same titrant solution.
- a) Justify the need to subtract the blank volume from the total volume measured.
- b) Calculate the mass of iodine ($\mathrm{I_2}$) present in the test sample. Data: $M(\mathrm{I}) = 126.9\,\mathrm{g\times mol^{-1}}$.
The sodium hypochlorite ($\mathrm{NaClO}$) content of a commercial bleaching product is analyzed by indirect iodometric titration. A $5.00\,\mathrm{mL}$ sample of bleach…
The sodium hypochlorite ($\mathrm{NaClO}$) content of a commercial bleaching product is analyzed by indirect iodometric titration. A $5.00\,\mathrm{mL}$ sample of bleach is diluted to $250.0\,\mathrm{mL}$ with distilled water. An aliquot of $20.00\,\mathrm{mL}$ of this diluted solution is acidified with acetic acid and an excess of potassium iodide is added, producing the quantitative reaction:
$\mathrm{ClO^-} + 2\,\mathrm{I^-} + 2\,\mathrm{H^+} \rightarrow \mathrm{I_2} + \mathrm{Cl^-} + \mathrm{H_2O}$
The iodine produced is titrated with sodium thiosulfate $0.0400\,\mathrm{mol\times L^{-1}}$, requiring $18.50\,\mathrm{mL}$ until the iodine-starch complex is decolorized:
$\mathrm{I_2} + 2\,\mathrm{S_2O_3^{2-}} \rightarrow 2\,\mathrm{I^-} + \mathrm{S_4O_6^{2-}}$
- a) Calculate the moles of sodium hypochlorite in the aliquot titrated.
- b) Determine the concentration of sodium hypochlorite in the original commercial solution in grams per liter. Data: atomic molar masses in $\mathrm{g\times mol^{-1}}$: $\mathrm{Na} = 23.0$; $\mathrm{Cl} = 35.5$; $\mathrm{O} = 16.0$.
Percent Purity and Concentration of a Commercial Sample
Store-bought products such as bleach or hydrogen peroxide are too concentrated to titrate directly. You dilute first, then titrate a portion (an aliquot).
Hydrogen peroxide acts as a reducing agent toward potassium permanganate in sulfuric acid medium. To determine the concentration of a commercial hydrogen peroxide…
Hydrogen peroxide acts as a reducing agent toward potassium permanganate in sulfuric acid medium. To determine the concentration of a commercial hydrogen peroxide solution whose density is $1,015\,\mathrm{g\times mL^{-1}}$, $2.00\,\mathrm{mL}$ of the sample is taken and diluted in a volumetric flask to $100.0\,\mathrm{mL}$. An aliquot of $20.00\,\mathrm{mL}$ of this solution is then titrated with $\mathrm{KMnO_4}$ $0.0150\,\mathrm{mol\times L^{-1}}$, using $19.60\,\mathrm{mL}$ of titrant.
Reaction: $2\,\mathrm{MnO_4^-} + 5\,\mathrm{H_2O_2} + 6\,\mathrm{H^+} \rightarrow 2\,\mathrm{Mn^{2+}} + 5\,\mathrm{O_2} + 8\,\mathrm{H_2O}$
- a) Calculate the molar concentration of $\mathrm{H_2O_2}$ in the commercial hydrogen peroxide solution.
- b) Determine the percentage by mass (% w/w) of $\mathrm{H_2O_2}$ in this commercial sample. Data: atomic molar masses in $\mathrm{g\times mol^{-1}}$: $\mathrm{H} = 1.0$; $\mathrm{O} = 16.0$.
Back Titrations and Blank Tests
Suppose you try to titrate a poorly soluble solid, or a slow reaction, directly. The drops fall but the indicator never shows a clear endpoint. How can you still measure the amount accurately?
The purity of a mineral containing pyrolusite ($\mathrm{MnO_2}$) is analyzed by a redox back titration. $0.2450\,\mathrm{g}$ of powdered mineral is weighed and treated…
The purity of a mineral containing pyrolusite ($\mathrm{MnO_2}$) is analyzed by a redox back titration. $0.2450\,\mathrm{g}$ of powdered mineral is weighed and treated while hot with $50.00\,\mathrm{mL}$ of a standard solution of sodium oxalate ($\mathrm{Na_2C_2O_4}$) $0.0500\,\mathrm{mol\times L^{-1}}$ in excess sulfuric acid medium, producing complete dissolution of the manganese dioxide according to the reaction:
$\mathrm{MnO_2} + \mathrm{C_2O_4^{2-}} + 4\,\mathrm{H^+} \rightarrow \mathrm{Mn^{2+}} + 2\,\mathrm{CO_2} + 2\,\mathrm{H_2O}$
The excess oxalate that did not react is titrated while hot at $60\,^\circ\mathrm{C}$ with $\mathrm{KMnO_4}$ $0.0200\,\mathrm{mol\times L^{-1}}$, using $17.85\,\mathrm{mL}$. A blank test of the auxiliary reagents, with the same acidic medium but without mineral or added oxalate, requires $0.35\,\mathrm{mL}$ of $\mathrm{KMnO_4}$.
Titration reaction: $2\,\mathrm{MnO_4^-} + 5\,\mathrm{C_2O_4^{2-}} + 16\,\mathrm{H^+} \rightarrow 2\,\mathrm{Mn^{2+}} + 10\,\mathrm{CO_2} + 8\,\mathrm{H_2O}$
- a) Calculate the net moles of oxalate that actually reacted with the manganese dioxide of the mineral.
- b) Determine the percentage by mass of $\mathrm{MnO_2}$ in the mineral analyzed. Data: atomic molar masses in $\mathrm{g\times mol^{-1}}$: $\mathrm{Mn} = 54.94$; $\mathrm{O} = 16.00$. Assume that the gangue is inert and that the blank consumption corresponds exclusively to reducing impurities in the auxiliary reagents.
In the synthesis of an inorganic product, a solution is obtained containing iodate ($\mathrm{IO_3^-}$) and iodide ($\mathrm{I^-}$) ions. When strongly acidified with…
In the synthesis of an inorganic product, a solution is obtained containing iodate ($\mathrm{IO_3^-}$) and iodide ($\mathrm{I^-}$) ions. When strongly acidified with sulfuric acid, comproportionation of both ions occurs, forming free molecular iodine ($\mathrm{I_2}$):
$\mathrm{IO_3^-} + 5\,\mathrm{I^-} + 6\,\mathrm{H^+} \rightarrow 3\,\mathrm{I_2} + 3\,\mathrm{H_2O}$
The starting point is a solution initially containing $1,200\,\mathrm{mmol}$ of $\mathrm{KIO_3}$ and an excess of $\mathrm{KI}$. After quantitatively collecting the iodine recovered from the process, it is titrated with sodium thiosulfate according to:
$\mathrm{I_2} + 2\,\mathrm{S_2O_3^{2-}} \rightarrow 2\,\mathrm{I^-} + \mathrm{S_4O_6^{2-}}$
Given that a solution of $\mathrm{Na_2S_2O_3}$ $0.1500\,\mathrm{mol\times L^{-1}}$ was used and that the net volume consumed was $43.20\,\mathrm{mL}$:
- a) Calculate the theoretical amount of $\mathrm{I_2}$ that should have formed, assuming a yield of $100\,\%$ in the comproportionation reaction.
- b) Determine the actual moles of $\mathrm{I_2}$ obtained from the titration and deduce the overall percentage yield of iodine production and recovery.
You have a pure salt containing only iron(II) oxalate dihydrate, of formula $\mathrm{FeC_2O_4 \times 2\,H_2O}$. In dilute, hot sulfuric acid medium, both constituent…
You have a pure salt containing only iron(II) oxalate dihydrate, of formula $\mathrm{FeC_2O_4 \times 2\,H_2O}$. In dilute, hot sulfuric acid medium, both constituent ions ($\mathrm{Fe^{2+}}$ and $\mathrm{C_2O_4^{2-}}$) are oxidized simultaneously by the permanganate ion:
$\mathrm{Fe^{2+}} \rightarrow \mathrm{Fe^{3+}} + \mathrm{e^-}$
$\mathrm{C_2O_4^{2-}} \rightarrow 2\,\mathrm{CO_2} + 2\,\mathrm{e^-}$
$\mathrm{MnO_4^-} + 8\,\mathrm{H^+} + 5\,\mathrm{e^-} \rightarrow \mathrm{Mn^{2+}} + 4\,\mathrm{H_2O}$
- a) Deduce the balanced overall ionic equation using the ion-electron method and determine the total number of electrons transferred per formula unit of $\mathrm{FeC_2O_4}$.
- b) If $0.2159\,\mathrm{g}$ of this pure salt is dissolved in excess sulfuric acid medium and titrated at $60\,^\circ\mathrm{C}$, determine the volume (in milliliters) of $\mathrm{KMnO_4}$ $0.0240\,\mathrm{mol\times L^{-1}}$ solution that will be used until the persistent pink end point is reached. Data: atomic molar masses in $\mathrm{g\times mol^{-1}}$: $\mathrm{Fe} = 55.85$; $\mathrm{C} = 12.01$; $\mathrm{O} = 16.00$; $\mathrm{H} = 1.01$.