(Will mark BRAINLIEST + 20 points - 5 for each question) * PLEASE DONT ANSWER IF U DONT KNOW* Barium Fluoride (BaF2) has a Ksp= 2.5 x 10^-5 (mol/L)^3

i. write the dissolution of BaF2 including all states:
ii. write the expression for Ksp for BaF2:
iii. write Ksp expression as an algebraic equation using the variable x to represent concentrations:
iv. find the solubility of BaF2 by solving the equation you wrote in part iii:

Silver bromide (AgBr) has a solubility of 7.07 × 10​–7​ mol/L. A. Write the dissolution. ... A. Write the dissolution reaction of BaF 2 , including all states.

Explanation:Molar solubility is the number of moles that are dissolved in 1 L solution.  

when BaF₂ dissolves it dissociates into the following ions

BaF₂ --> Ba²⁺ + 2F⁻

if the molar solubility of BaF₂ is X, then molar solubility of Ba²⁺ is X and F⁻ is 2x

then the formula for the solubility product constant -ksp is;

ksp = [Ba²⁺][F⁻]²

ksp = X * (2X)²

ksp = 4X³

since ksp = 1.7 x 10⁻⁶

4X³ = 1.7 x 10⁻⁶

X = 0.0075 M

molar solubility of BaF₂ is 0.0075m

thats how its supposed to be

yes

ya te ayud espera