Problem 26 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The volume (in \(\mathrm{mL}\) ) of \(0.1 \mathrm{~N} \mathrm{NaOH}\) required to neutralise \(10 \mathrm{~mL}\) of 0.1 \(\mathrm{N}\) phosphinic acid is $$ \text { [Main Sep. } \mathbf{0 3}, \mathbf{2 0 2 0}[\mathbf{I I})] $$

Step-by-Step Solution

Answer

20 mL of 0.1 N NaOH is required.

Step 1: Understanding the Problem

The problem asks for the volume of a sodium hydroxide solution required to neutralize a phosphinic acid solution. Both solutions have a concentration of 0.1N.

Step 2: Write the Neutralization Reaction

The chemical reaction for neutralizing phosphinic acid (also known as hypophosphorous acid, H₃PO₂) with sodium hydroxide (NaOH) can be written as: \[\text{H}_3\text{PO}_2 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{HPO}_2 + 2\text{H}_2\text{O}\].This shows that 2 moles of NaOH are required to neutralize 1 mole of phosphinic acid.

Step 3: Determine Equivalent Milliequivalents

The equivalent milliequivalents (meq) of H₃PO₂ is given by: \[\text{meq of H}_3\text{PO}_2 = 10 \text{ mL} \times 0.1 \text{ N} = 1\]meq because phosphinic acid donates 1 acidic hydrogen to the reaction.

Step 4: Calculate Required meq of NaOH

Since 2 meq of NaOH are needed to neutralize 1 meq of H₃PO₂:\[\text{meq of NaOH required} = 2 \times 1 = 2\].

Step 5: Calculate Volume of NaOH Solution

Using the formula: \[\text{Volume (mL)} = \frac{\text{meq}}{\text{Normality}}\]we find the volume of NaOH needed:\[\text{Volume of NaOH} = \frac{2 \text{ meq}}{0.1 \text{ N}} = 20 \text{ mL}\].

Key Concepts

Understanding Stoichiometry

In chemistry, stoichiometry is the method used to determine the quantities of reactants and products in chemical reactions.
It involves the calculation of mass, volume, and concentrations based on the balanced chemical equations of reactions. Calculating stoichiometric ratios requires understanding how many moles of each substance are involved.
In this exercise, stoichiometry tells us how sodium hydroxide (NaOH) and phosphinic acid (H₃PO₂) interact during the neutralization.

• The balanced chemical equation \[ ext{H}_3 ext{PO}_2 + 2 ext{NaOH} ightarrow ext{Na}_2 ext{HPO}_2 + 2 ext{H}_2 ext{O} \] explains that two moles of NaOH are required to neutralize each mole of phosphinic acid.
• This ratio is crucial for calculating the exact amounts needed in the reaction.
• Thus, stoichiometry helps in ensuring that the reaction proceeds completely without excess of any reactant.

Exploring Molarity and Normality

Molarity is a measure of the concentration of a solute in a solution, expressed as moles per liter.
In this solution, it is essential to recognize the relationship between molarity and normality, especially while working with acids and bases. For phosphinic acid,

• The concentration provided (\(0.1 ext{ N}\)) indicates that it donates one acidic hydrogen, aligning with its role in the reaction.
• Similarly, NaOH, also with a normality of \(0.1 ext{ N}\), reflects its strength as a base in this reaction.

To calculate the volume needed for neutralization, normality simplifies since normality directly correlates with the equivalent moles needed.
Instead of calculating molarity separately, normality already considers the ion exchange specific to acid-base reactions. This simplicity aids in solving problems quickly, as it bridges the specific reaction behavior directly to the calculated values used in the equations.

Balancing Chemical Equations

Chemical equations represent the symbolic depiction of chemical reactions where reactants transform into products.
They need to be balanced to reflect the conservation of mass, where atom counts for each element are equal on both sides of the equation. In our original equation with phosphinic acid and NaOH,

• Each element's atom count is crucial; therefore, understanding that two NaOH molecules react with one phosphinic acid is determined from balancing.
• It ensures that no atoms are lost or gained, highlighting the fundamental law that mass is conserved in reactions.

Balancing chemical equations helps in calculating the exact amount of each substance required.
In a real-world lab scenario, ensuring equations are balanced is essential for your calculated reactants to result in desired products without wastage, excess, or shortage.