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Chapter 17: Problem 7

Write short notes on: (i) Iodimetry (ii) Iodometry (iii) Kjeldahl's process

Short Answer

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Iodimetry involves direct titration of iodine; iodometry involves indirect titration by iodine; Kjeldahl's process measures nitrogen content.

Step by step solution

01

Understanding Iodimetry

Iodimetry is a type of volumetric chemical analysis, a method used to determine the concentration of a substance. It involves the direct titration of iodine (I鈧�) with a reducing agent, usually a thiosulfate salt ( \[ ext{2S}_2 ext{O}_3^{2-} \] ), which reduces the iodine to iodide ions ( \[ ext{I}^- \] ). This method is typically used for substances that can react with, and thus detect, free iodine directly. A starch solution is often used as an indicator, which forms a deep blue complex with iodine, signifying the endpoint of the titration when the blue color disappears.
02

Understanding Iodometry

Iodometry is another form of titrimetric method similar to iodimetry but involves the indirect measurement of substances that can reduce iodine. In iodometry, an analyte reduces iodine to iodide ions, and the excess iodine produced is titrated with a standard solution of sodium thiosulfate. The amount of titrant used provides an indirect measure of the analyte. Like iodimetry, a starch indicator is used to determine the endpoint by the disappearance of the blue-black starch-iodine complex.
03

Understanding Kjeldahl's Process

Kjeldahl's process is a widely used method in analytical chemistry to quantify the nitrogen content in organic compounds. The process involves three main steps: digestion, neutralization, and titration. In digestion, a sample is digested with concentrated sulfuric acid, which converts nitrogen in organic material into ammonium sulfate. Following digestion, the solution is neutralized with a strong base, releasing ammonia. Finally, ammonia is distilled into a solution containing a known volume of standardized acid, and the amount of excess acid is determined by back titration. This determines the original nitrogen content in the sample.

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Key Concepts

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

Iodimetry
Iodimetry is a fascinating technique in analytical chemistry, predominantly used when you need to determine the concentration of certain substances. At the heart of this method is the titration of iodine (\( \text{I}_2 \)) with a reducing agent often provided by a solution of thiosulfate ions (\( \text{2S}_2\text{O}_3^{2-} \)).
  • The process begins with the reagent, iodine, which acts as an oxidizing agent.
  • It reacts with specific compounds, and is then reduced to iodide ions (\( \text{I}^- \)) by the thiosulfate.
  • A starch solution is often used as an indicator. This is because it creates a deep blue color when combined with iodine, which disappears at the titration's endpoint.

This color change signals the reaction's completion, allowing chemists to measure the substance's concentration accurately. This method is particularly helpful for substances that can bind directly with iodine.
Iodometry
Iodometry is another intricate titrimetric analysis method, closely related to iodimetry, but with some key differences. It involves the indirect analysis of a substance's concentration by measuring how much iodine is reduced by the analyte itself.
  • An unknown substance or analyte reduces iodine (\( \text{I}_2 \)) to iodide (\( \text{I}^- \)).
  • The excess iodine remaining after this reaction is then titrated with a standard solution of sodium thiosulfate.
  • Similar to iodimetry, iodometry uses a starch indicator that forms a blue-black color with iodine, marking the endpoint when the color disappears.

Iodometry is especially useful when working with oxidizing agents. By determining the volume of thiosulfate used, you can calculate the amount of the original analyte, making it an indirect measurement method.
Kjeldahl's Process
Kjeldahl's process is a classic and essential technique in analytical chemistry. It's primarily employed to measure the nitrogen content in organic compounds, which has applications in agriculture, food science, and more.
  • The procedure starts with digestion, where the sample is treated with concentrated sulfuric acid. This stage converts organic nitrogen into ammonium sulfate.
  • Next comes neutralization, where the acidic solution is neutralized with a strong base, releasing ammonia (\( \text{NH}_3 \)).
  • Finally, the released ammonia is distilled and trapped in a known volume of standardized acid. The excess acid is then determined by back titration.

This multi-step process helps in uncovering the original nitrogen content within a sample. It's renowned for its accuracy and reliability in measuring nitrogen, which is crucial in understanding and managing agricultural practices and food nutrients.

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Most popular questions from this chapter

\(4 \mathrm{~g}\) of a mixture of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) and anhydrous \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) were dissolved in pure water and volume made up to \(250 \mathrm{~mL} .20\) \(\mathrm{mL}\) of this solution required \(25 \mathrm{~mL}\) of \(\mathrm{N} / 5 \mathrm{H}_{2} \mathrm{SO}_{4}\) for complete neutralisation. Calculate the percentage composition of the mixture.

\(3.45 \mathrm{~g}\) of a metallic carbonate were mixed with \(240 \mathrm{~mL}\) of \(N / 4 \mathrm{HCl}\). The excess acid was neutralised by \(50 \mathrm{~mL}\) of \(\mathrm{N} / 5\) KOH solution. Calculate the equivalent mass of the metal. [Hint : Equivalent mass of metal carbonate $$ =\frac{3.45 \times 4 \times 1000}{200}=69 $$ Equivalent mass of metal \(=69-\) eq. mass of carbonate $$ =(69-30)=39] $$

One litre of a mixture of \(\mathrm{O}_{2}\) and \(\mathrm{O}_{3}\) at NTP was allowed to react with an excess of acidified solution of KI. The iodine liberated required \(40 \mathrm{~mL}\) of \(M / 10\) sodium thiosulphate solution for titration. What is the weight per cent of ozone in the mixture? Ultraviolet radiation of wavelength \(300 \mathrm{~nm}\) can decompose ozone. Assuming that one photon can decompose one ozone molecule, how many photons would have been required for the complete decomposition of ozone in the original mixture? \(\quad\) [I.L.T. 1997] [Hint : Let the total moles of \(\mathrm{O}_{2}\) and \(\mathrm{O}_{3}\) in the mixture be \(n .\) Applying $$ P V=n R T $$ $$ \begin{aligned} 1 \times 1 &=n \times 0.0821 \times 273 \\ n &=0.044 \text { mole } \end{aligned} $$ Moles of \(\mathrm{O}_{3}=\) moles of \(\mathrm{I}_{2}\) $$ \begin{aligned} &=1 / 2 \text { moles of } \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \\ &=\frac{1}{2} \times \frac{1}{10} \times \frac{40}{1000} \\ &=0.002 \mathrm{~mole} \end{aligned} $$ Moles of \(\mathrm{O}_{2}\) in the mixture \(=0.044-0.002=0.042 \mathrm{~mole}\) Mass of \(\mathrm{O}_{2}=0.042 \times 32=1.344 \mathrm{~g}\) $$ \begin{aligned} \text { Mass of } \mathrm{O}_{3} &=0.002 \times 48=0.096 \mathrm{~g} \\ \% \mathrm{O}_{3} &=\frac{0.096}{1.44} \times 100=6.6 \end{aligned} $$ No. of photons required to decompose \(0.002\) mole of ozone $$ \left.=0.002 \times 6.02 \times 10^{23}=1.204 \times 10^{21}\right]^{-} $$

\(1.6 \mathrm{~g}\) of pyrolusite was treated with \(60 \mathrm{~mL}\) of normal oxalic acid and some \(\mathrm{H}_{2} \mathrm{SO}_{4}\). The oxalic acid left undecomposed was made upto \(250 \mathrm{~mL}, 25 \mathrm{~mL}\) of this solution required \(32 \mathrm{~mL}\) of \(0.1 \mathrm{~N}\) potassium permanganate \(\left(\mathrm{KMnO}_{4}\right) .\) Calculate the percentage of pure \(\mathrm{MnO}_{2}\) in pyrolusite.

What is the weight in grams of available \(\mathrm{O}_{2}\) per litre from a solution of \(\mathrm{H}_{2} \mathrm{O}_{2}, 10 \mathrm{~mL}\) of which when titrated with \(\mathrm{N} / 20\) \(\mathrm{KMnO}_{4}\) solution required \(25 \mathrm{~mL}\) for the reaction, $$ \begin{aligned} 2 \mathrm{KMnO}_{4}+5 \mathrm{H}_{2} \mathrm{O}_{2}+& 4 \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \\ & 5 \mathrm{O}_{2}+8 \mathrm{H}_{2} \mathrm{O} \end{aligned}+2 \mathrm{KHSO}_{4}+2 \mathrm{MnSO}_{4} $$
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