Calculate equivalent weight of K2Cr2O7 in acidic medium.

To calculate the equivalent weight of potassium dichromate (K2Cr2O7) in an acidic medium, we first need to understand what equivalent weight means in this context. The equivalent weight of a compound is defined as the mass of that compound that will combine with or displace one mole of hydrogen ions (H+) in a reaction. In acidic conditions, K2Cr2O7 acts as an oxidizing agent, particularly in redox reactions.

Understanding the Reaction

In acidic medium, potassium dichromate is reduced from Cr(VI) to Cr(III). The balanced half-reaction for this reduction can be represented as follows:

• Cr2O7^2- + 14 H⁺ + 6 e^- → 2 Cr³⁺ + 7 H₂O

This equation shows that for every mole of K2Cr2O7, which contains two chromium atoms, six electrons are transferred during the reduction process. Therefore, the number of equivalents of K2Cr2O7 can be determined based on the number of electrons transferred.

Calculating Molar Mass

Next, we need to calculate the molar mass of K2Cr2O7:

• Potassium (K): 39.1 g/mol × 2 = 78.2 g/mol
• Chromium (Cr): 52.0 g/mol × 2 = 104.0 g/mol
• Oxygen (O): 16.0 g/mol × 7 = 112.0 g/mol

Adding these together gives:

Molar mass of K2Cr2O7 = 78.2 + 104.0 + 112.0 = 294.2 g/mol

Determining Equivalent Weight

Now that we have the molar mass, we can calculate the equivalent weight. The formula for equivalent weight is:

Equivalent Weight = Molar Mass / n

Where n is the number of electrons transferred per mole of the substance in the reaction. For K2Cr2O7, n = 6 (as derived from the half-reaction).

Substituting the values we have:

Equivalent Weight = 294.2 g/mol / 6 = 49.03 g/equiv

Final Thoughts

Thus, the equivalent weight of K2Cr2O7 in acidic medium is approximately 49.03 g/equiv. This value is crucial for stoichiometric calculations in redox reactions where K2Cr2O7 is used as an oxidizing agent. Understanding this concept helps in various applications, such as titrations and analytical chemistry, where precise measurements are essential.