Question

Barium fluoride (BaF2) has a Ksp = 2. 5 × 10–5 (mol/L)3.

Answer 1

This problem is asking for the dissolution reaction of barium fluoride, both the equilibrium and Ksp expressions in terms of concentrations and x and its molar solubility in water. Thus, answers shown below:

• $BaF_2(s)\rightleftharpoons Ba^{2+}(aq)+2F^-(aq)$
• $Ksp=[Ba^{2+}][F^-]^2$
• $Ksp=(x)(2x)^2$
• 0.0184 M.

Solubility product

In chemistry, when a solid is dissolved in water, one must take into account the fact that not necessarily its 100 % will be able to break into ions and thus undergo dissolution.

In such a way, and specially for sparingly soluble solids, one ought to write the dissolution reaction at equilibrium as shown below for the given barium fluoride:

$BaF_2(s)\rightleftharpoons Ba^{2+}(aq)+2F^-(aq)$

Next, we can write its equilibrium expression according to the law of mass action, which also demands us to omit any solid and refer it to the solubility product constant (Ksp):

$Ksp=[Ba^{2+}][F^-]^2$

Afterwards, one can insert the reaction extent, x, as it stands for the molar solubility of this solid in water, taking into account the coefficients balancing the reaction:

$Ksp=(x)(2x)^2$

Finally, we solve for the x as the molar solubility of barium fluoride as shown below:

$2.5x10^{-5}=(x)(2x)^2\\\\2.5x10^{-5}=4x^3\\\\x=\sqrt[3]{\frac{2.5x10^{-5}}{4} } \\\\x=0.0184M$

Learn more about chemical equilibrium: brainly.com/question/26453983