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: (a) \(10 \mathrm{~mL}\) (b) \(40 \mathrm{~mL}\) (c) \(60 \mathrm{~mL}\) (d) \(80 \mathrm{~mL}\)

Short answer

The required volume of 0.1 M KOH is 40 mL.

Step-by-step solution

Analyze the Reaction

Phosphorus acid, H₃PO₃, is a diprotic acid, which means it can donate two protons (H⁺) in reactions. The neutralization reaction with KOH, a strong base, will occur as follows: \[ \text{H}_3\text{PO}_3 + 2\text{KOH} \rightarrow \text{K}_2\text{HPO}_3 + 2\text{H}_2\text{O} \]This shows that each molecule of H₃PO₃ requires 2 molecules of KOH for complete neutralization.

Calculate Moles of H₃PO₃

The moles of H₃PO₃ present in the solution can be calculated using the formula: \[ \text{Moles} = Molarity \times \text{Volume (L)} \]Given the molarity is 0.1 M and the volume is 20 mL (which is 0.020 L), we calculate:\[ \text{Moles of H₃PO₃} = 0.1 \times 0.020 = 0.002 \text{ moles} \]

Calculate Moles of KOH Required

Since the reaction requires 2 moles of KOH per mole of H₃PO₃, we need to multiply the moles of H₃PO₃ by 2 to find the moles of KOH required:\[ \text{Moles of KOH} = 2 \times 0.002 = 0.004 \text{ moles} \]

Calculate Volume of KOH Solution

The volume of KOH solution needed can be found using the formula: \[ \text{Volume (L)} = \frac{\text{Moles}}{\text{Molarity}} \]Given the molarity of KOH is 0.1 M, we calculate:\[ \text{Volume of KOH} = \frac{0.004}{0.1} = 0.040 \text{ L} = 40 \text{ mL} \]

Conclusion

After completing the calculations, we conclude that 40 mL of 0.1 M KOH solution is required to completely neutralize 20 mL of 0.1 M H₃PO₃ solution.

Key concepts

Phosphorus Acid

Phosphorus acid (chemical formula: H₃PO₃) is a diprotic acid, which means it has the ability to donate two protons (H⁺ ions). This property characterizes it as having two stages in its neutralization process where it reacts with bases. In the context of chemistry, understanding whether an acid is monoprotic, diprotic, or triprotic is important. This distinction impacts how many equivalents of a base are required for complete neutralization.

For the neutralization of phosphorus acid, since it's diprotic, it will need two moles of a monoprotic base such as KOH for complete reaction. This reaction significanty informs the stoichiometry of acid-base reactions, which is crucial when performing chemical calculations. Keep this in mind when calculating how much base is needed for reactions involving H₃PO₃.

Molarity Calculations

Molarity is a key concept in chemistry, representing the concentration of a solution. It's denoted in units of moles per liter (mol/L or M). Understanding this concept is crucial for performing accurate chemical calculations and predicting reaction outcomes.

When given the volume and molarity of a solution, you can easily find the number of moles of a solute using the formula:

• Moles = Molarity × Volume (L)

For example, in the original exercise, we calculated the moles of H₃PO₃ from a 20 mL (0.020 L) solution with a molarity of 0.1 M to be 0.002 moles. This calculation forms the basis for finding out how much reactant is needed for complete neutralization.

Chemical Reaction Stoichiometry

In chemical reactions, stoichiometry is the quantitative relationship between reactants and products in a balanced chemical equation. It is essential for predicting how much reactant(s) is needed or how much product will be formed.

Stoichiometry helps us determine the amount of each substance involved in a reaction, such as in the neutralization of phosphorus acid by KOH. In the given equation:

• H₃PO₃ + 2KOH → K₂HPO₃ + 2H₂O

This tells us that 1 mole of H₃PO₃ requires 2 moles of KOH for completion. Thus, these ratios directly inform how to calculate the amount of KOH needed to neutralize a given quantity of H₃PO₃ in a solution. Consequently, it assists in determining that 40 mL of the KOH solution is required for the complete neutralization of the specified amount of phosphorus acid.