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(a) \(2 \mathrm{~g}\) of a metal carbonate were dissolved in \(50 \mathrm{~mL}\) of \(N\) HCl. \(100 \mathrm{~mL}\) of \(0.1 \mathrm{~N} \mathrm{NaOH}\) were required to neutralise the resultant solution. Calculate the equivalent mass of the metal carbonate. (b) How much water should be added to \(75 \mathrm{~mL}\) of \(3 \mathrm{NHCl}\) to make it a normal solution?

Short Answer

Expert verified
(a) Equivalent mass of metal carbonate is 0.05 g/meq. (b) Add 150 mL of water.

Step by step solution

01

Calculate Initial Acid Milliequivalents

First, calculate the milliequivalents of HCl used initially. Since 1 normal solution means 1 equivalent per liter, you can calculate the milliequivalents using:\[\text{Milliequivalents of } \text{HCl} = \text{Volume (mL)} \times \text{Normality} = 50 \text{ mL} \times 1 = 50 \text{ meq}\]
02

Calculate Base Milliequivalents

Calculate the milliequivalents of NaOH used for neutralization. Using the same formula for milliequivalents, we get:\[\text{Milliequivalents of } \text{NaOH} = \text{Volume (mL)} \times \text{Normality} = 100 \text{ mL} \times 0.1 = 10 \text{ meq}\]
03

Determine Acid Consumed by Metal Carbonate

The milliequivalents of HCl consumed by the metal carbonate can be determined by subtracting the milliequivalents of NaOH from the initial acid milliequivalents.\[\text{Milliequivalents of HCl consumed by metal carbonate} = 50 \text{ meq} - 10 \text{ meq} = 40 \text{ meq}\]
04

Calculate Equivalent Mass of Metal Carbonate

The equivalent mass of the metal carbonate is calculated by dividing the mass of the metal carbonate by the milliequivalents of acid consumed. \[\text{Equivalent mass} = \frac{\text{Mass of metal carbonate}}{\text{Milliequivalents of HCl consumed}} = \frac{2 \text{ g}}{40} = 0.05 \text{ g/meq}\]
05

Calculate Water Needed for Dilution

Now, calculate how much water needs to be added to 75 mL of 3N HCl to make it a normal (1N) solution. Use the dilution formula:\[C_1V_1 = C_2V_2\]where \(C_1 = 3N\), \(V_1 = 75\text{ mL}\), \(C_2 = 1N\), and \(V_2\) is the final volume. Solve for \(V_2\):\[3 \times 75 = 1 \times V_2 \225 = V_2\]So the final volume should be 225 mL. Thus, the amount of water to add is:\[V_2 - V_1 = 225 \text{ mL} - 75 \text{ mL} = 150 \text{ mL}\]

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Key Concepts

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

Neutralization
Neutralization is an essential concept in chemistry that happens when an acid reacts with a base to form water and a salt. In the reaction, the hydrogen ions ( H^+ ) from the acid combine with the hydroxide ions ( OH^- ) from the base, neutralizing their effects. For example, in our problem, hydrochloric acid (HCl) reacts with sodium hydroxide (NaOH) to achieve neutralization.
This is indicated by the fact that the acid leftover from the initial solution is neutralized by the added base. In the given exercise, understanding neutralization allows us to calculate how much of the acid reacted with the metal carbonate versus how much was neutralized by the NaOH. This gives us the critical information needed to determine the equivalent mass of the metal carbonate.
Normality Calculations
Normality is a measure of concentration equivalent to molarity but considers the reactive capacity of a compound. This unit is often used in acid-base reactions where equivalents of reactive ions are central. Unlike molarity, normality accounts for the change in concentration of ions based on their participation in chemical reactions.
To calculate normality, the formula used is n_{eq} = V imes N , where V is the volume in milliliters, and N denotes normality. In the given problem, the milliequivalents of HCl and NaOH are determined by multiplying their volumes by their respective normalities. Through these calculations, we gain insight into the chemical stoichiometry of the reactions being considered.
Dilution Process
The dilution process involves reducing the concentration of a solute in a solution, usually by adding more solvent, such as water. This concept is critical when adjusting the normality of a solution to suit particular experimental requirements. In our problem, we deal with diluting a concentrated 3N HCl solution to make it a normal (1N) solution.
The key calculation for dilution uses the formula C_1V_1 = C_2V_2 , which ensures the amount of solute remains constant before and after dilution. By adjusting our volumes in this way, we calculated the final volume required and, consequently, the volume of water needed for dilution.
Molarity versus Normality
Molarity and normality are two different ways to express a solution's concentration. Molarity (M) is used to specify the number of moles of solute per liter of solution and provides information about the total number of solute particles. In contrast, normality (N) considers equivalent concentration, which reflects the solution's potential reaction capacity.
Normality can be more versatile, as it involves the actual participation rate of ions in reactions—something molarity doesn't account for. For example, sulfuric acid ( H_2SO_4 ) has a molarity of 1M, but its normality could be 2N, reflecting its dual hydrogen ions, each capable of reacting. This distinction is particularly relevant in reactions like neutralization, requiring precise stoichiometry control based on reactive agents.

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Most popular questions from this chapter

\(1.64 \mathrm{~g}\) of a mixture of calcium carbonate and magnesium carbonate were dissolved in \(50 \mathrm{~mL}\) of \(0.8 \mathrm{~N}\) hydrochloric acid. The excess of the acid required \(16 \mathrm{~mL} \mathrm{~N} / 4\) sodium hydroxide solution for neutralisation. Find out the percentage composition of the mixture of two carbonates.

6 g sample of \(\left(\mathrm{Fe}_{3} \mathrm{O}_{4}+\mathrm{Fe}_{2} \mathrm{O}_{3}\right)\) mixture and an inert material was treated with excess of aqueous \(\mathrm{KI}\) which reduces all \(\mathrm{Fe}^{3+}\) to \(\mathrm{Fe}^{2+}\) ion. The resulting solution was diluted to \(50 \mathrm{~mL}\). \(10 \mathrm{~mL}\) of the diluted solution titrated with \(5.5 \mathrm{~mL}\) of \(1 \mathrm{M}\) \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) to react with all iodine. The \(\mathrm{I}_{2}\) from another \(25 \mathrm{~mL}\) sample was extracted after which the \(\mathrm{Fe}^{2+}\) was titrated with \(3.2 \mathrm{~mL}\) of \(1 \mathrm{MMnO}_{4}^{-}\) in \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution. Calculate percentage of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) and \(\mathrm{Fe}_{3} \mathrm{O}_{4}\) in the mixture. [Hint : Reactions involved are : (1) \(\mathrm{Fe}_{3} \mathrm{O}_{4}+2 \mathrm{KI}+4 \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow 3 \mathrm{FeSO}_{4}+\mathrm{K}_{2} \mathrm{SO}_{4}+4 \mathrm{H}_{2} \mathrm{O}+\mathrm{I}_{2}\) (2) \(\mathrm{Fe}_{2} \mathrm{O}_{3}+2 \mathrm{KI}+3 \mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow 2 \mathrm{FeSO}_{4}+\mathrm{K}_{2} \mathrm{SO}_{4}+3 \mathrm{H}_{2} \mathrm{O}+\mathrm{I}_{2}\) (3) \(\mathrm{I}_{2}+2 \mathrm{~S}_{2} \mathrm{O}_{3}^{2-} \longrightarrow 2 \Gamma+\mathrm{S}_{4} \mathrm{O}_{6}^{2-}\) (4) \(\left.\mathrm{MnO}_{4}^{-}+5 \mathrm{Fe}^{2+}+8 \mathrm{H}^{+} \longrightarrow \mathrm{Mn}^{2+}+5 \mathrm{Fe}^{3+}+4 \mathrm{H}_{2} \mathrm{O}\right]\). \(25 \mathrm{~m} \mathrm{~L} \mathrm{H}_{2} \mathrm{O}_{2}\) were added to excess of acidified solution of \(\mathrm{KI}\). The iodine so liberated required \(20 \mathrm{~mL}\) of \(0.1 \mathrm{~N}\) sodium

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 $$ \begin{aligned} &=\frac{0.01 \times 6.02 \times 10^{23}}{1000} \times 100 \\ &\left.=6.02 \times 10^{20}\right] \end{aligned} $$

A \(1.2\) g mixture of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) and \(\mathrm{K}_{2} \mathrm{CO}_{3}\) was dissolved in water to form \(100 \mathrm{~cm}^{3}\) of a solution. \(20 \mathrm{~cm}^{3}\) of this solution required \(40 \mathrm{~cm}^{3}\) of \(0.1 \mathrm{~N} \cdot \mathrm{HCl}\) for neutralisation. \(\mathrm{Calculate}\) the mass of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) and \(\mathrm{K}_{2} \mathrm{CO}_{3}\) in the mixture.

What do you mean by molarity and normality of the solution? How are these related ?

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