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Chapter 21: Problem 43

The equilibrium constant at \(25^{\circ} \mathrm{C}\) for the reaction $$\mathrm{Br}_{2}(l)+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{Br}^{-}(a q)+\mathrm{HBrO}(a q)$$ is \(1.2 \times 10^{-9} .\) This is the system present in a bottle of "bromine water." Assuming that HBrO does not ionize appreciably, what is the pH of the bromine water?

Short Answer

Expert verified
Answer: The pH of the bromine water is approximately 2.98.

Step by step solution

01

Write the balanced equation and equilibrium expression

The balanced equation is: $$\mathrm{Br}_{2}(l)+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{Br}^{-}(a q)+\mathrm{HBrO}(a q)$$ The equilibrium expression is: $$K = \frac{[\mathrm{H}^{+}][\mathrm{Br}^{-}][\mathrm{HBrO}]}{[\mathrm{Br}_{2}][\mathrm{H}_{2}\mathrm{O}]}$$ However, since the concentration of liquid substances (like \(\mathrm{Br}_{2}\) and \(\mathrm{H}_{2}\mathrm{O}\)) do not affect the equilibrium constant, we can rewrite the expression as: $$K = [\mathrm{H}^{+}][\mathrm{Br}^{-}][\mathrm{HBrO}]$$
02

Set up the ICE table

We begin by setting up an ICE table to find the equilibrium concentrations of each species: |Species |Initial |Change |Equilibrium| |-----------|--------|-------|-----------| |H+ | 0 |+x |x | |Br− | 0 |+x |x | |HBrO | 0 |+x |x | Since we are assuming that HBrO does not ionize significantly, the equilibrium concentrations of H+, Br- and HBrO are all equal to x.
03

Substitute the equilibrium concentrations into the equilibrium expression

Substituting the equilibrium concentrations into the equilibrium expression, we get: $$K = (x)(x)(x) = x^3$$ As we know, the value of K is \(1.2 \times 10^{-9}\). So, we have: $$x^3 = 1.2 \times 10^{-9}$$
04

Solve for x (concentration of H+ ions)

To find the concentration of H+ ions, we need to solve for x: $$x = \sqrt[3]{1.2 \times 10^{-9}}$$ Now we can use a calculator to find the value of x: $$x \approx 1.06 \times 10^{-3}$$ This represents the concentration of H+ ions at equilibrium.
05

Calculate the pH of the solution

Now that we know the concentration of H+ ions, we can use the pH formula to find the pH of the bromine water: $$pH = -\log_{10}([\mathrm{H}^{+}])$$ Substitute the value of x into the formula: $$pH = -\log_{10}(1.06 \times 10^{-3})$$ Now we can use a calculator to find the pH: $$pH \approx 2.98$$ The pH of the bromine water is approximately 2.98.

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Most popular questions from this chapter

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 \(\mathrm{HF}(a q)\).

Write a balanced net ionic equation for the disproportionation reaction (a) of iodine to give iodate and iodide ions in basic solution. (b) of chlorine gas to chloride and perchlorate ions in basic solution.

Write a balanced net ionic equation for (a) the electrolytic decomposition of hydrogen fluoride. (b) the oxidation of iodide ion to iodine by hydrogen peroxide in acidic solution. Hydrogen peroxide is reduced to water.

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

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