Question

the value of the solubility product constant for silver bromide is
write the reaction that corresponds to this ksp value.
ksp values are found by clicking on the \tables\ link.
be sure to specify the state of each reactant or product.
if a box is not needed leave it blank.

Explanation:

Step1: Recall the solubility equilibrium of silver bromide

Silver bromide (AgBr) is a sparingly soluble salt. In water, it dissociates into silver ions ($\text{Ag}^+$) and bromide ions ($\text{Br}^-$) according to the equilibrium reaction: $\text{AgBr}(s)
ightleftharpoons \text{Ag}^+(aq) + \text{Br}^-(aq)$

Step2: Determine the $K_{sp}$ expression and reaction

The solubility product constant ($K_{sp}$) for this reaction is given by $K_{sp} = [\text{Ag}^+][\text{Br}^-]$. The reaction corresponding to the $K_{sp}$ of AgBr is the dissociation of solid AgBr into its aqueous ions.

Answer:

The reaction is $\boldsymbol{\text{AgBr}(s)
ightleftharpoons \text{Ag}^+(aq) + \text{Br}^-(aq)}$ and the $K_{sp}$ value for silver bromide is $5.3 \times 10^{-13}$ (this is the known $K_{sp}$ value for AgBr at 25°C; if you were to calculate it from solubility, you would use the equilibrium concentrations, but the standard $K_{sp}$ for AgBr is $5.3 \times 10^{-13}$). For the reaction part, filling in the boxes: first box $\text{AgBr}(s)$, second box $\text{Ag}^+(aq)$, third box $\text{Br}^-(aq)$ (note: the format with + signs is for the reactant and products, so the reaction is $\text{AgBr}(s)
ightarrow \text{Ag}^+(aq) + \text{Br}^-(aq)$ but in equilibrium form as above).