use the chemical equation to answer the question.\n\n$co(h_{2}o)_{6}^{2+} + 4cl^{-} + heat…

use the chemical equation to answer the question.\n\n$co(h_{2}o)_{6}^{2+} + 4cl^{-} + heat \\rightleftharpoons cocl_{4}^{2-} + 6h_{2}o$\n\ncobalt ions form complex ions with water and chloride. the $co(h_{2}o)_{6}^{2+}$ ions are pink, and the $cocl_{4}^{2-}$ ions are blue. a scientist creates a solution that is purple at equilibrium. how can the scientist make the solution more pink?\n\n(1 point)\n\n- by adding water $h_{2}o$\n- by adding heat to the system\n- by adding hydrochloric acid (hcl)\n- by adding $co(h_{2}o)_{6}^{2+}$ ions

use the chemical equation to answer the question.\n\n$co(h_{2}o)_{6}^{2+} + 4cl^{-} + heat \\rightleftharpoons cocl_{4}^{2-} + 6h_{2}o$\n\ncobalt ions form complex ions with water and chloride. the $co(h_{2}o)_{6}^{2+}$ ions are pink, and the $cocl_{4}^{2-}$ ions are blue. a scientist creates a solution that is purple at equilibrium. how can the scientist make the solution more pink?\n\n(1 point)\n\n- by adding water $h_{2}o$\n- by adding heat to the system\n- by adding hydrochloric acid (hcl)\n- by adding $co(h_{2}o)_{6}^{2+}$ ions

Answer

Explanation:

Step1: Identify the target color and ion

The pink color is associated with the reactant $\text{Co}(\text{H}{2}\text{O}){6}^{2+}$. To make the solution more pink, the equilibrium must shift to the left.

Step2: Apply Le Chatelier's Principle to products

According to Le Chatelier's Principle, increasing the concentration of a product will shift the equilibrium toward the reactants (left).

Step3: Evaluate the effect of adding water

Water ($\text{H}{2}\text{O}$) is a product in the given chemical equation: $$\text{Co}(\text{H}{2}\text{O}){6}^{2+} + 4\text{Cl}^{-} + \text{heat} \rightleftharpoons \text{CoCl}{4}^{2-} + 6\text{H}{2}\text{O}$$ Adding $\text{H}{2}\text{O}$ increases the product concentration, shifting the equilibrium to the left, thereby increasing the concentration of the pink $\text{Co}(\text{H}{2}\text{O}){6}^{2+}$ ions.

Answer:

by adding water $\text{H}_{2}\text{O}$