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Chapter 21: Problem 47

What is the concentration of fluoride ion in a water solution saturated with \(\mathrm{BaF}_{2}, K_{s \mathrm{p}}=1.8 \times 10^{-7}\) ?

Short Answer

Expert verified
Answer: The concentration of fluoride ions in the saturated water solution is approximately \(5.88 \times 10^{-3}\,\text{M}\).

Step by step solution

01

Write the dissolution equilibrium equation

Write the balanced chemical equation for the dissolution of BaF2 in water: \[\text{BaF}_2(s) \rightleftharpoons \text{Ba}^{2+}(aq) + 2\text{F}^-(aq)\]
02

Write the solubility product expression

Write the expression for Ksp for BaF2: \[K_\text{sp} = [\text{Ba}^{2+}][\text{F}^-]^2\]
03

Define variables for ion concentrations

Let x be the concentration of Ba2+ ions at equilibrium. Then, the concentration of F- ions at equilibrium will be 2x because there are two F- ions for each Ba ion: \[[\text{Ba}^{2+}] = x\] \[[\text{F}^-] = 2x\]
04

Substitute concentrations into the Ksp expression and solve for x

Insert the expressions for the ion concentrations into the Ksp expression, and then solve for x: \[K_\text{sp} = [\text{Ba}^{2+}][\text{F}^-]^2 = (x)(2x)^2\] \[K_\text{sp} = 4x^3\] \[x = \sqrt[3]{\frac{K_\text{sp}}{4}}\]
05

Calculate the value of x

Substitute the given value for Ksp into the equation and solve for x: \[x = \sqrt[3]{\frac{1.8 \times 10^{-7}}{4}}\] \[x ≈ 2.94 \times 10^{-3}\,\text{M}\]
06

Calculate the concentration of fluoride ions

Now that we have the concentration of Ba2+ ions at equilibrium (x), we can find the concentration of F- ions by multiplying x by 2: \[[\text{F}^-] = 2x = 2 \times 2.94 \times 10^{-3}\,\text{M} \approx 5.88 \times 10^{-3}\,\text{M}\] In conclusion, the concentration of fluoride ions in the saturated water solution is approximately \(5.88 \times 10^{-3}\,\text{M}\).

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Most popular questions from this chapter

Complete and balance the following equations. If no reaction occurs, write NR. (a) \(\mathrm{Cl}_{2}(g)+\mathrm{I}^{-}(a q) \longrightarrow\) (b) \(\mathrm{F}_{2}(g)+\mathrm{Br}^{-}(a q) \longrightarrow\) (c) \(\mathrm{I}_{2}(s)+\mathrm{Cl}^{-}(a q) \longrightarrow\) (d) \(\mathrm{Br}_{2}(l)+\mathrm{I}^{-}(a q) \longrightarrow\)

Write the formula for the following compounds. (a) sodium azide (b) sulfurous acid (c) hydrazine (d) sodium dihydrogen phosphate

Consider the equilibrium system $$\mathrm{HF}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{F}^{-}(a q)$$ Given \(\Delta H_{\mathrm{f}}^{\circ} \mathrm{HF}(a q)=-320.1 \mathrm{~kJ} / \mathrm{mol}\), $$\begin{aligned} \Delta H_{\mathrm{f}}^{\circ} \mathrm{F}^{-}(a q) &=-332.6 \mathrm{~kJ} / \mathrm{mol} ; S^{\circ} \mathrm{F}^{-}(a q)=-13.8 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} \\ K_{\mathrm{a}} \mathrm{HF} &=6.9 \times 10^{-4} \text { at } 25^{\circ} \mathrm{C} \end{aligned}$$ calculate \(S^{\circ}\) for \(\mathrm{HF}(a q)\).

Give the Lewis structure of (a) \(\mathrm{Cl}_{2} \mathrm{O}\) (b) \(\mathrm{N}_{2} \mathrm{O}\) (c) \(\mathrm{P}_{4}\) (d) \(\mathbf{N}_{2}\)

Explain why (a) acid strength increases as the oxidation number of the central non- metal atom increases. (b) nitrogen dioxide is paramagnetic. (c) the oxidizing strength of an oxoanion is inversely related to \(\mathrm{pH}\). (d) sugar turns black when treated with concentrated sulfuric acid.
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