Question

To 100.0 mL of a solution that contains 0.120 M Cr(NO₃)₂ and 0.500 M HNO₃ is added 20.0 mL of 0.250 M K₂Cr₂O₇. The dichromate and chromium (II) ions react to give chromium (III) ions.

(a) Write a balanced net ionic equation for the reaction.
(b) Calculate the concentrations of all ions in the solution after reaction.

Answer 1

The net ionic equation for the reaction has been provided, and the concentrations of the ions after reaction can be determined by taking into account the volume change and applying stoichiometry to calculate the moles reacted and remaining.

Step-by-step explanation:

To start with, we need to write the net ionic equation for the reaction between Cr(NO3)2 and K2Cr2O7 where dichromate and chromium (II) ions react to produce chromium (III) ions.

Cr²⁺⁺ (aq) + Cr₂O₇²⁺ (aq) + 14H⁺ (aq) → 2Cr³⁺ (aq) + 7H₂O (l)

The initial concentration of Cr(NO3)2 is 0.120 M in 100.0 mL and the concentration of HNO3 is 0.500 M. The concentration of K2Cr2O7 is 0.250 M in 20.0 mL. To find the concentration of all ions after the reaction, we must account for the volume change and use the stoichiometry of the reaction.

Since the dichromate ion is reduced and the chromium(II) ion is oxidized, we use stoichiometry to calculate the moles reacted and the remaining. After the reaction, we'll use the final volume of the solution (100.0 mL + 20.0 mL = 120.0 mL) to calculate the final concentrations of the ions.