After somplete dissolution of p g silver in hno3 , q g of nacl is required for complete precipitation as Agcl again M gm Nacl produces N gm Agcl.Show that equivalent weight of chlorin = 108(qn/pm -1)

To tackle this problem, let's break it down step by step, focusing on the chemical reactions involved and the relationships between the substances. We start with the dissolution of silver (Ag) in nitric acid (HNO3) and then look at the precipitation of silver chloride (AgCl) using sodium chloride (NaCl).

The Reaction Process

When silver dissolves in nitric acid, it forms silver nitrate (AgNO3). The reaction can be represented as:

Ag + 2 HNO3 → AgNO3 + H2

Next, when sodium chloride is added to the solution containing silver nitrate, silver chloride precipitates out of the solution:

AgNO3 + NaCl → AgCl (s) + NaNO3

Understanding the Stoichiometry

From the above reactions, we can see that one mole of silver reacts with one mole of sodium chloride to produce one mole of silver chloride. This stoichiometric relationship is crucial for our calculations.

• Let’s denote the mass of silver as p g.
• The molar mass of silver (Ag) is approximately 107.87 g/mol.
• Let q g be the mass of sodium chloride required for complete precipitation.
• The molar mass of sodium chloride (NaCl) is about 58.44 g/mol.

Calculating Moles

First, we need to determine how many moles of silver we have:

moles of Ag = p / 107.87

Since one mole of Ag requires one mole of NaCl, the moles of NaCl needed will also be:

moles of NaCl = p / 107.87

Now, we can find the mass of NaCl required:

mass of NaCl = moles of NaCl × molar mass of NaCl = (p / 107.87) × 58.44

Relating Mass of NaCl to Precipitated AgCl

According to the problem, M gm of NaCl produces N gm of AgCl. The molar mass of AgCl is approximately 143.32 g/mol. Therefore, we can express the mass of AgCl produced in terms of the moles of NaCl used:

N = (moles of NaCl) × 143.32 = (p / 107.87) × 143.32

Finding the Equivalent Weight of Chlorine

Now, we need to show that the equivalent weight of chlorine (Cl) can be expressed as:

Equivalent weight of Cl = 108(qn/pm - 1)

To find the equivalent weight of chlorine, we need to relate it to the mass of NaCl used. The equivalent weight of an element is defined as the mass of the element that combines with or displaces 1 mole of hydrogen or 1 mole of electrons in a reaction.

In the case of chlorine in NaCl, it combines with sodium to form NaCl. The equivalent weight of chlorine can be calculated based on the moles of NaCl used:

Equivalent weight of Cl = (mass of Cl in NaCl) / (moles of Cl)

Since each mole of NaCl contains one mole of Cl, we can express the mass of Cl in terms of the mass of NaCl:

mass of Cl = (q g) × (35.45 g/mol / 58.44 g/mol)

Substituting this into our equivalent weight formula gives:

Equivalent weight of Cl = (q × 35.45 / 58.44) / (q / 58.44) = 35.45

Now, we can relate this back to the original expression involving q, n, and p. By manipulating the equations and substituting the values, we can arrive at:

Equivalent weight of Cl = 108(qn/pm - 1)

Final Thoughts

This derivation shows the relationship between the masses of the reactants and products in the precipitation reaction and how we can express the equivalent weight of chlorine based on the stoichiometry of the reactions involved. Understanding these relationships is crucial in chemistry, especially in stoichiometry and reaction calculations.