Question

To neutralize completely \(20 \mathrm{~mL}\) of \(0.1 \mathrm{M}\) aqueous solution of phosphorus acid, the volume of \(0.1 \mathrm{M}\) aqueous KOH solution required is [2004] (a) \(10 \mathrm{~mL}\) (b) \(40 \mathrm{~mL}\) (c) \(60 \mathrm{~mL}\) (d) \(80 \mathrm{~mL}\)

Short answer

(b) 40 mL

Step-by-step solution

Understand Phosphorus Acid Neutralization

Phosphorus acid is often represented as H₃PO₃. It can donate two protons (H⁺) because it is a diprotic acid, indicated by the number of replaceable hydrogen ions for each formula unit. In neutralization, KOH will provide one hydroxide ion (OH⁻) per mole.

Calculate Moles of Phosphorus Acid

We have 20 mL (or 0.020 L) of 0.1 M phosphorus acid. The moles of H₃PO₃ are calculated by multiplying the molarity by the volume:\[moles ext{ } of ext{ } H_3PO_3 = 0.1 ext{ } M imes 0.020 ext{ } L = 0.002 ext{ } moles\]

Determine Moles of KOH Required

Since H₃PO₃ can donate two protons and requires two OH⁻ ions to be fully neutralized, we need two moles of KOH per mole of H₃PO₃:\[moles ext{ } of ext{ } KOH = 2 imes 0.002 ext{ } moles = 0.004 ext{ } moles\]

Calculate Volume of KOH Solution

Using the equation for molarity, the volume ( V ) of KOH solution needed is:\[moles ext{ } of ext{ } KOH = M imes V ightarrow V = \frac{0.004 ext{ } moles}{0.1 ext{ } M} = 0.040 ext{ } L = 40 ext{ } mL\]

Conclusion

The volume of 0.1 M KOH solution required to completely neutralize the 20 mL of 0.1 M phosphorus acid solution is 40 mL.

Key concepts

Understanding Phosphorus Acid (H₃PO₃)

Phosphorus acid is chemically known as H₃PO₃, a compound that plays a significant role in various chemical reactions. In the world of chemistry, phosphorus acid is unique due to its composition and behavior in reactions.
First off, it's important to note that phosphorus acid is a diprotic acid. This means it has the ability to donate two protons (H⁺ ions) during its chemical reactions, specifically in neutralization processes.
Why is this important? Well, this trait determines how phosphorus acid interacts with bases like KOH. Since it's diprotic, it requires two molecules of a base like KOH to completely neutralize one molecule of H₃PO₃.
In practical scenarios, understanding these characteristics guides scientists and students alike in predicting and calculating the outcomes of neutralization reactions, such as the exercise described.

Characteristics of Diprotic Acids

Diprotic acids, like phosphorus acid, have two replaceable hydrogen ions. In chemical reactions:

• These acids will first lose one hydrogen ion, and then the second, in two distinct steps.
• Each step involves the donation of a proton, which is crucial in the context of neutralization reactions with bases.

A valuable concept when working with diprotic acids is the stoichiometry of their reactions. Since they release protons in two stages, the calculations for neutralization involve ensuring enough base is present to react with both protons each acid molecule releases. In the specific exercise, 20 mL of 0.1 M phosphorus acid contains 0.002 moles of the acid, which effectively means that 0.004 moles of base (KOH) are needed, due to its diprotic nature, to ensure complete neutralization.

Steps in Molarity Calculation

Molarity plays a crucial role in understanding chemical reactions, especially those involving acids and bases. It is defined as the number of moles of solute per liter of solution. Here, we are focusing on how to calculate how much KOH is needed to neutralize phosphorus acid in the specified scenario.
Let's break down the steps:

• First, calculate the moles of phosphorus acid present: Multiply the volume of the acid (in liters) by its molarity.
• Next, apply stoichiometry to find out how many moles of KOH are needed, considering phosphorus acid's diprotic nature.
• Finally, use the formula for molarity to find the volume of KOH solution needed: Rearrange the formula to solve for volume (V = moles of solute / molarity).

In the problem, 0.004 moles of KOH are necessary to completely neutralize the phosphorus acid. Given the molarity of KOH is 0.1 M, the calculation ends up requiring 40 mL of KOH, which showcases the practical application of molarity in real-world chemical problems. This step-by-step calculation forms the backbone of successfully solving molarity-related questions.