Question

write the dissolution reaction for manganese(ii) bromide in water.
be sure to specify the state of each reactant and product.
is manganese(ii) bromide considered soluble or not soluble?
a. soluble
b. not soluble
based upon this, the equilibrium constant for this reaction will be:
a. greater than 1
b. less than 1

Explanation:

Part 1: Dissolution Reaction of Manganese(II) Bromide

Step 1: Identify the compound and its ions

Manganese(II) bromide has the formula $\ce{MnBr2}$. In water, it dissociates into $\ce{Mn^2+}$ and $\ce{Br^-}$ ions.

Step 2: Write the dissolution reaction

The solid compound dissolves to form aqueous ions:
$$\ce{MnBr2(s) -> Mn^2+(aq) + 2Br^-(aq)}$$

Part 2: Solubility of Manganese(II) Bromide

Manganese(II) bromide is a salt of a Group 2 metal (manganese is a transition metal, but bromides of most transition metals, including Mn²⁺, are soluble in water, especially when paired with bromide ions (Br⁻), as bromides are generally soluble except for a few exceptions (e.g., AgBr, PbBr₂, Hg₂Br₂). MnBr₂ is soluble.

For a dissolution reaction $\ce{MnBr2(s) <=> Mn^2+(aq) + 2Br^-(aq)}$, the equilibrium constant ($K_{sp}$) is given by $K_{sp} = [\ce{Mn^2+}][\ce{Br^-}]^2$. Since $\ce{MnBr2}$ is soluble, the concentration of ions in solution is significant, meaning the products (ions) are favored. For soluble salts, $K_{sp}$ (or the equilibrium constant for dissolution) is typically greater than 1 (as more product is formed at equilibrium).

Answer:

A. Soluble

Part 3: Equilibrium Constant for the Reaction