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If \(1.26 \mathrm{~g}\) of oxalic acid is dissolved in \(250 \mathrm{~mL}\) of solution, find its normality. The equivalent mass of oxalic acid is 63 .

Short Answer

Expert verified
The normality of the oxalic acid solution is 0.08 N.

Step by step solution

01

Understand Normality Formula

The normality ( extit{N}) of a solution is calculated using the formula: \[ N = \frac{\text{Weight of solute (g)}}{\text{Eq. mass of solute (g/equiv)}} \times \frac{1}{\text{Volume of solution (L)}} \] Given: Weight of oxalic acid = 1.26 g, Eq. mass of oxalic acid = 63 g/equiv, Volume of solution = 250 mL = 0.250 L.
02

Substitute Values into the Formula

Substitute the given values into the normality formula: \[ N = \frac{1.26}{63} \times \frac{1}{0.250} \] Calculate the fraction and multiply by the inverse of the volume in liters.
03

Calculate the Normality

First, calculate the division of the weight by the equivalent mass: \[ \frac{1.26}{63} \approx 0.02 \] Then multiply this result by the inverse of the volume: \[ 0.02 \times \frac{1}{0.250} = 0.02 \times 4 = 0.08 \] Thus, the normality of the solution is 0.08 N.

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Key Concepts

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

Equivalent Mass
Equivalent mass is a crucial concept in chemistry, especially when dealing with titrations and normality calculations. It is defined as the mass of a substance that will react with or replace one mole of hydrogen ions (+). For ionizable oxalic acid in our exercise, the equivalent mass is given as 63 g/equiv.

To find the equivalent mass, you need to know the molar mass of the compound and its valency or change capacity. The formula to calculate it is:
  • Equivalent Mass = \( \frac{\text{Molar Mass}}{\text{Valency}} \)
In simpler terms, you divide the molar mass by the number of equivalents the compound can furnish. For oxalic acid, it can donate two moles of hydrogen ions, making its valency 2. This simplification helps in calculations because it reduces the amount of substance calculations to the equivalent level needed for reaction balance.
Oxalic Acid
Oxalic acid is an organic compound with the formula \(\text{C}_2\text{H}_2\text{O}_4\), and it features prominently in various chemical reactions due to its strong acidic nature. It is a dicarboxylic acid, meaning it contains two carboxyl functional groups (\(-\text{COOH}\)), contributing to its ability to donate protons in reactions.

In its pure form, oxalic acid appears as a colorless, crystalline compound and is widely used in industries such as textile, metal cleaning, and bleaching. It naturally occurs in many plants like rhubarb and spinach. When determining its concentration in solutions, understanding its behavior is essential, especially since it has an equivalent mass that simplifies its involvement in stoichiometric calculations. This measure ensures precise computations in laboratory and industrial chemical applications.
Chemical Calculations
Chemical calculations in the context of the given exercise focus on determining normality, which involves a specific method for measuring concentration based on the equivalent concept.

Normality is closely related to the molarity of a solution but adds a dimension of chemical reactivity with the use of equivalents. It is expressed using the formula:
  • Normality \( (N) = \frac{\text{Weight of solute (g)}}{\text{Equivalent mass of solute (g/equiv)}} \times \frac{1}{\text{Volume of solution (L)}} \)
In our exercise, we see step-by-step how oxalic acid’s normality is calculated by substituting the weight, equivalent mass, and volume in the given formula.

In essence, these calculations allow chemists to determine not just how many molecules there are per liter (like molarity), but how many reactive entities there are, which is crucial for reactions where one equivalent equals one mole of reactive or charge-altering species. Whether preparing solutions or conducting titrations, understanding chemical calculations using normality provides a more fine-tuned measurement for engaging with chemical reactions.

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Most popular questions from this chapter

In an ore the only oxidisable material is \(\mathrm{Sn}^{2+} .\) This ore is titrated with a dichromate solution containing \(2.5 \mathrm{~g} \mathrm{~K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) in \(0.50\) litre. A \(0.40 \mathrm{~g}\) of sample of the ore required \(10.0 \mathrm{~cm}^{3}\). of the titrant to reach equivalent point. Calculate the percentage of tin in ore. \((\mathrm{K}=39.1, \mathrm{Cr}=52, \mathrm{Sn}=118.7)\) [Roorkee 1993] [Hint : Mol. mass of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}=2 \times 39.1+2 \times 52+7 \times 16\) \(=78.2+104.0+112.0\) \(=294.2\) Eq. mass of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}=\frac{294.2}{6}=49.03\) Normality of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) solution \(=\frac{2.5}{49.03} \times \frac{1000}{500}=\frac{5}{49.03} N\) \(10 \mathrm{~mL} \frac{5}{49.03} \mathrm{~N} \mathrm{~K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \equiv 10 \mathrm{~mL} \frac{5}{49.03} \mathrm{~N}\) stannous ion Eq. mass of \(\mathrm{Sn}^{2+}=\frac{118.7}{2}=59.35\) Amount of Sn in the sample \(=\frac{5}{49.03} \times \frac{59.35}{1000} \times 10\) \(=0.0605 \mathrm{~g}\) Percentage of \(\mathrm{Sn}\) in the ore \(\left.=\frac{0.0605}{0.40} \times 100=15.125\right]\)

A sample of hard water contains \(96 \mathrm{ppm}\) of \(\mathrm{SO}_{4}^{2-}\) and 183 ppm of \(\mathrm{HCO}_{3}^{-}\), with \(\mathrm{Ca}^{2+}\) as the only cation. How many moles of \(\mathrm{CaO}\) will be required to remove \(\mathrm{HCO}_{3}{ }^{-}\) from \(1000 \mathrm{~kg}\) of this water? If \(1000 \mathrm{~kg}\) of this water is treated with the amount of \(\mathrm{CaO}\) calculated above, what will be the concentration (in ppm) of residual \(\mathrm{Ca}^{2+}\) ions? (Assume \(\mathrm{CaCO}_{3}\) to be completely insoluble in water). If the \(\mathrm{Ca}^{2+}\) ions in one litre of the treated water are completely exchanged with hydrogen ions, what will be its \(\mathrm{pH}\) ? (One ppm means one part of the substance in one million parts of water.)

Calculate the number of oxalic acid molecules in \(100 \mathrm{~mL}\) of \(\begin{array}{ll}0.02 N \text { oxalic acid solution. } & \text { [Roorkee 1991] }\end{array}\) [Hint : Molarity \(=\frac{0.02}{2}=0.01 M\) No. of molecules in one molar solution \(=6.02 \times 10^{23}\) No. of molecules in \(100 \mathrm{~mL}\) of \(0.01 M\) oxalic acid solution $$ \begin{aligned} &=\frac{0.01 \times 6.02 \times 10^{23}}{1000} \times 100 \\ &\left.=6.02 \times 10^{20}\right] \end{aligned} $$

\(10 \mathrm{~g}\) of a commercial sample of copper sulphate were dissolved in water and the volume made up to \(500 \mathrm{~mL} .25\) \(\mathrm{mL}\) of this solution was treated with excess of \(\mathrm{KI}\) and the liberated iodine required \(30 \mathrm{~mL}\) of \(\mathrm{N} / \mathrm{l} 2 \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) solution. Calculate the percentage of copper in the sample.

\(50 \mathrm{~mL}\) of \(0.2 \mathrm{~N} \mathrm{KMnO}_{4}\) is required for complete oxidation of \(0.45 \mathrm{~g}\) of anhydrous oxalic acid. Calculate the normality of oxalic acid solution.

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