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A quantity of \(\mathrm{KMnO}_{4}\) was boiled with \(\mathrm{HCl}\) and the gas evolved was led into a solution of \(\mathrm{KI}\). When the reaction was complete, the \(\mathrm{I}_{2}\) liberated was titrated with a solution of hypocontaining \(124 \mathrm{~g}\) of \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) per litre. It was found that exactly \(60 \mathrm{~mL}\) were required to decolourise the solution of \(\mathrm{I}_{2}\). What weight of \(\mathrm{KMnO}_{4}\) was used?

Short Answer

Expert verified
0.948 grams of \( \mathrm{KMnO}_{4} \) were used.

Step by step solution

01

Calculate Molar Mass of Na2S2O3路5H2O

First, find the molar mass of sodium thiosulphate pentahydrate, Na鈧係鈧侽鈧兟5H鈧侽. - Na: 23 g/mol, so for 2 atoms: 46 g/mol - S: 32 g/mol, so for 2 atoms: 64 g/mol - O: 16 g/mol, so for 3 atoms: 48 g/mol - 5 H鈧侽: 5 x (2*1 + 16) = 90 g/mol Add these values together: 46 + 64 + 48 + 90 = 248 g/mol.
02

Determine Moles of Na2S2O3路5H2O in Solution

Next, calculate the number of moles of Na鈧係鈧侽鈧兟5H鈧侽 in 60 mL of solution. - The solution concentration is 124 g/L, equivalent to 124 g in 1000 mL. - Convert grams to moles, using the molar mass: \( \text{moles} = \frac{124}{248} = 0.5 \text{ moles per litre} \).- Find moles in 60 mL: \( \text{moles} = 0.5 \times \frac{60}{1000} = 0.03 \text{ moles} \).
03

Relate I2 and Na2S2O3

When I鈧 is titrated with Na鈧係鈧侽鈧, the reaction is \( \mathrm{I}_{2} + 2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-} \rightarrow 2\mathrm{I}^{-} + \mathrm{S}_{4} \mathrm{O}_{6}^{2-} \). This indicates a 1:2 molar ratio. Therefore, 0.03 moles of Na鈧係鈧侽鈧 correspond to 0.015 moles of I鈧.
04

Relate KMnO4 and I2

Understand the reaction of \( \mathrm{KMnO}_{4} \) with \( \mathrm{I}^{-} \) in acidic solution:\( 2\mathrm{KMnO}_{4} + 16\mathrm{HCl} + 10\mathrm{KI} \rightarrow 2\mathrm{MnCl}_{2} + 5\mathrm{I}_{2} + 8\mathrm{H}_{2}\mathrm{O} + 2\mathrm{KCl} \).This reaction shows 2 moles of \( \mathrm{KMnO}_{4} \) produce 5 moles of \( \mathrm{I}_{2} \).Therefore, if 0.015 moles of \( \mathrm{I}_{2} \) were produced, 0.006 moles of \( \mathrm{KMnO}_{4} \) were involved (\( 0.015 \times \frac{2}{5} \)).
05

Calculate Mass of KMnO4

Finally, determine the mass of \( \mathrm{KMnO}_{4} \) used.- The molar mass of \( \mathrm{KMnO}_{4} \) is: K (39 g/mol) + Mn (55 g/mol) + 4 x O (4 x 16 g/mol) = 158 g/mol.- Calculate the mass: mass = moles x molar mass: \( \text{mass} = 0.006 \times 158 = 0.948 \text{ grams} \).Thus, approximately 0.948 grams of \( \mathrm{KMnO}_{4} \) were used.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Potassium Permanganate
Potassium permanganate (KMnO鈧) is a powerful oxidizing agent widely used in redox reactions, which involve the transfer of electrons between species. Its formula consists of one potassium (K), one manganese (Mn), and four oxygen (O) atoms, giving it a molar mass of 158 g/mol. In redox processes, potassium permanganate undergoes reduction, often changing from a purple color (due to Mn in the +7 oxidation state) to a colorless or light pink color (as Mn is reduced to the +2 state in an acidic environment). This visual change makes it particularly useful in titration setups, allowing analysts to determine the endpoint of a reaction visually by the disappearance or appearance of color.

In the specific context of the given exercise, potassium permanganate is reacted with hydrochloric acid (HCl) and a solution of potassium iodide (KI). The reaction sequence that ensues is well-charted in redox chemistry. The MnO鈧勨伝 ions are reduced, while iodide ions (I鈦) are oxidized to iodine (I鈧), which is subsequently titrated. Understanding this progression is crucial for grasping the quantitative aspects of chemical reactions, as it defines how much of each reactant is necessary for a complete reaction, and how much product will be formed.
Sodium Thiosulfate
Sodium thiosulfate (Na鈧係鈧侽鈧兟5H鈧侽) is a critical reagent, especially in iodine titrations due to its ability to react quantitatively with iodine. In the oxidation state where thiosulfate is present, it acts as a mild reducing agent and is frequently used to titrate iodine (I鈧) in chemical analyses.

The role of sodium thiosulfate in titration is to reduce iodine back to iodide ions. This specific reaction is defined by the chemical equation: \[ \mathrm{I}_{2} + 2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-} \rightarrow 2\mathrm{I}^{-} + \mathrm{S}_{4} \mathrm{O}_{6}^{2-} \] This equation highlights a 1:2 molar ratio between iodine and thiosulfate. Understanding this stoichiometry is essential for determining how much sodium thiosulfate would be required to fully react with a certain amount of iodine present in the solution.

In the exercise example, the concentration of sodium thiosulfate is provided, allowing us to calculate the number of moles used in the titration process. This calculation is a stepping stone to determining the amount of iodine present initially, leading us back to the quantity of KMnO鈧 that was used.
Iodine Titration
Iodine titration is a technique that involves using iodine as the titrating agent, commonly for determining the amount of a reducing agent present in a sample. The procedure often requires an accurate indication of when all iodine has reacted, which is typically shown by a color change facilitated by a starch indicator, forming a blue complex with iodine that disappears when iodine is fully reacted.

In the context of this exercise, the titration process determines the iodine produced from a previous redox reaction, where iodide ions were oxidized by the action of potassium permanganate. By knowing the stoichiometry of the reaction between iodine and sodium thiosulfate, we can precisely calculate the moles of iodine titrated. This, in turn, reveals the initial amount of KMnO鈧 responsible for the oxidation.

Understanding iodine titration is crucial for many applications beyond this exercise. It's widely used in various fields, such as pharmaceuticals, cosmetics, and even in food testing, to ascertain the concentration of oxidizing or reducing agents. Iodine's distinct color and chemistry play a pivotal role in its functionality in these processes.

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Most popular questions from this chapter

\(30 \mathrm{~mL}\) of sodium carbonate solution is mixed with \(20 \mathrm{~mL}\) of \(0.8 \mathrm{~N}\) sulphuric acid. The resultant solution needed \(20 \mathrm{~mL}\) of \(0.7 \mathrm{~N}\) hydrochloric acid solution for complete neutralisation. Determine the strength of the sodium carbonate in grams per litre. (Take sodium carbonate to be anhydrous).

What will be the normality of solution obtained by mixing 20 \(\mathrm{mL}\) of \(0.5 \mathrm{~N}\) hydrochloric acid, \(60 \mathrm{~mL}\) of \(0.1 \mathrm{~N}\) sulphuric acid and \(150 \mathrm{~mL}\) of \(0.2 \mathrm{~N}\) nitric acid?

How many millilitres of \(0.5 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) are needed to dissolve \(\begin{array}{ll}0.5 \mathrm{~g} \text { of copper (II) carbonate? } & \text { [I.I.T. 1999] }\end{array}\) [Hint : \(N_{1} V_{1}=N_{2} V_{2}\) \(N_{1}=\) Normality of \(\mathrm{H}_{2} \mathrm{SO}_{4}=0.5 \times 2=1 \mathrm{~N}\) \(V_{1}=\) Vol. of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) \(N_{2}=\) Normality of copper (II) carbonate \(=\frac{0.5 \times 2}{123.5} \mathrm{~N}\) \(V_{2}=\) Volume of copper (II) carbonate \(=1000 \mathrm{~mL}\) Thus, \(1.0 \times V_{1}=\frac{0.5 \times 2}{123.5} \times 1000\) or \(\left.\quad V_{1}=8.09 \mathrm{~mL}\right]\)

A sample of magnesium was burnt in air to give mixture of \(\mathrm{MgO}\) and \(\mathrm{Mg}_{3} \mathrm{~N}_{2}\). The ash was dissolved in 60 meq. of \(\mathrm{HCl}\) and the resulting solution back titrated with \(\mathrm{NaOH} .12 \mathrm{meq}\). of \(\mathrm{NaOH}\) were required to reach the end point. An excess of \(\mathrm{NaOH}\) was then added and the solution distilled. The ammonia released was then trapped in 10 meq. of second acid solution. Back titration of this acid solution required 6 meq. of the base. Calculate percentage magnesium burnt to the nitride. \(\quad\) [Roorkee 1998] [Hint : Let moles of \(\mathrm{Mg}\) used for \(\mathrm{MgO}\) and \(\mathrm{Mg}_{3} \mathrm{~N}_{2}\) are \(x\) and \(\mathrm{y}\) respectively. \(2 \mathrm{Mg}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{MgO}\) \(3 \mathrm{Mg}+\mathrm{N}_{2} \longrightarrow \mathrm{Mg}_{3} \mathrm{~N}_{2}\) Moles of \(\mathrm{MgO}=x\), Moles of \(\mathrm{Mg}_{3} \mathrm{~N}_{2}=y / 3\) \(\mathrm{MgO}+2 \mathrm{HCl} \longrightarrow \mathrm{MgCl}_{2}+\mathrm{H}_{2} \mathrm{O}\) \(\mathrm{Mg}_{3} \mathrm{~N}_{2}+8 \mathrm{HCl} \longrightarrow 3 \mathrm{MgCl}_{2}+2 \mathrm{NH}_{4} \mathrm{Cl}\) tion will contain \(x\) moles of \(\mathrm{MgCl}_{2}\) and \(2 y / 3\) moles of \(\mathrm{NH}_{4} \mathrm{Cl}\). Moles of \(\mathrm{HCl}\) used \(=2 x+8 y / 3\) for \(\mathrm{MgO}\) and \(\mathrm{Mg}_{3} \mathrm{~N}_{2}\) \(2 x+\frac{8 y}{3}=60-12=48 \quad \ldots\) (i) Moles of \(\mathrm{NH}_{4} \mathrm{Cl}\) formed \(=\) moles of \(\mathrm{NH}_{3}\) \(=\) moles of HCl used to absorb \(\mathrm{NH}_{3}\) \(2 y / 3=4\) \(y=6\) From eq. (i) \(2 x+\frac{8 \times 6}{3}=48\) 5 \(\% \mathrm{Mg}\) used for \(\mathrm{Mg}_{3} \mathrm{~N}_{2}\) Page 1

\(1.0 \mathrm{~g}\) carbonate of a metal was dissolved in \(50 \mathrm{~mL} \mathrm{~N} / 2 \mathrm{HCl}\) solution. The resulting liquid required \(25 \mathrm{~mL}\) of \(\mathrm{N} / \mathrm{S} \mathrm{NaOH}\) solution to neutralise it completely. Calculate the equivalent mass of the metal carbonate.

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