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Normality of \(0.3 \mathrm{M} \mathrm{H}_{3} \mathrm{PO}_{4}\) solution is (a) \(0.1 \mathrm{~N}\) (b) \(0.45 \mathrm{~N}\) (c) \(0.6 \mathrm{~N}\) (d) \(0.9 \mathrm{~N}\)

Short Answer

Expert verified
The normality of the solution is \(0.9 \mathrm{~N}\), so the answer is (d).

Step by step solution

01

Understand the Concept of Normality

Normality (\(N\)) is a measure of concentration equivalent to molarity but accounts for the equivalent factor of the solute. For acids and bases, it refers to the number of moles of reactive ion provided per liter of solution.
02

Determine the Equivalent Factor

Phosphoric acid (\( \mathrm{H}_3\mathrm{PO}_4 \)) is triprotic, meaning it can donate three protons (\(\mathrm{H}^+\) ions) per molecule, so its equivalent factor is 3.
03

Use the Formula for Normality

The formula for normality is given by:\[ N = M \times f \]where \( M \) is the molarity and \( f \) is the equivalent factor. Here, \( M = 0.3 \text{ M} \) and \( f = 3 \).
04

Calculate the Normality

Substitute the given values into the formula:\[ N = 0.3 \times 3 = 0.9 \text{ N} \]
05

Provide the Answer

Based on the calculation, the normality of the \(0.3 \mathrm{M} \mathrm{H}_{3}\mathrm{PO}_{4}\) solution is \(0.9 \mathrm{~N}\). Therefore, the correct option is (d).

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Key Concepts

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

Normality calculation
Normality is a term frequently used in chemistry to describe the concentration of a solution. It's akin to molarity but delves a bit deeper by considering the equivalent factor of the solute. This makes it extra useful for reactions where the nature of the solute's activity matters.
If you think about an acid-base reaction, the normality would tell us about the "reactive potential" of an acid or base in that reaction. Essentially, it informs us about how many moles of reactant could react.
The formula to calculate normality is:
  • \( N = M \times f \)
  • Where \( N \) is the normality, \( M \) is the molarity, and \( f \) is the equivalent factor.
To find the normality of our solution of phosphoric acid (H_3PO_4), we'd utilize this formula together with molarity and the equivalent factor.
Equivalent factor
The concept of equivalent factor revolves around the reactive units of a substance engaging in a chemical reaction. For phosphoric acid ( H_3PO_4 ), it is critical to recognize it as a triprotic acid, meaning each molecule has the potential to donate three protons (H⁺ ions).
In chemical reactions, these protons play a pivotal role. Hence, the equivalent factor for phosphoric acid is 3, reflecting its ability to give up three protons.
To determine the correct equivalent factor:
  • For acids, it's the number of protons the acid can donate.
  • For bases, it's the number of hydroxide ions ( OH^- ) they can accept.
This specificity ensures that the normality value reflects the true chemical potential of the solution for the reaction being considered.
Phosphoric acid concentration
When considering the concentration of phosphoric acid (H_3PO_4), it's important to differentiate between molarity and normality. Molarity tells you the number of moles of solute per liter of solution, while normality incorporates the reactive capacity of each molecule.
Suppose you have a 0.3 M solution of phosphoric acid. To find its normality, you must factor in its equivalent factor. Since the equivalent factor is 3 (because it's triprotic), you multiply the molarity by this factor to get the normality:
  • Given: \( M = 0.3 \)
  • Equivalent factor \( f = 3 \)
Using the formula \( N = M \times f \), the normality would be:
  • \( N = 0.3 \times 3 = 0.9 \)
Thus, your phosphoric acid solution at a molarity of 0.3 has a normality of 0.9 N.

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Most popular questions from this chapter

Ratio of \(\mathrm{C}_{\mathrm{p}}\) and \(\mathrm{C}_{\mathrm{v}}\) of a gas ' \(\mathrm{X}^{\prime}\) is \(1.4\). The number of atoms of the gas ' \(\mathrm{X}\) ' present in \(11.2\) litres of it at NTP will be (a) \(6.02 \times 10^{21}\) (b) \(60.2 \times 10^{23}\) (c) \(6.02 \times 10^{23}\) (d) \(1.02 \times 10^{23}\)

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The equivalent weight of phosphoric acid \(\left(\mathrm{H}_{3} \mathrm{PO}_{4}\right)\) in the reaction: \(\mathrm{NaOH}+\mathrm{H}_{3} \mathrm{PO}_{4} \longrightarrow \mathrm{NaH}_{2} \mathrm{PO}_{4}+\mathrm{H}_{2} \mathrm{O}\) is (a) 89 (b) 98 (c) 59 (d) 29

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