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Chapter 16: Problem 72

Calculate the molar concentration of \(\mathrm{OH}^{-}\) in a \(0.724 \mathrm{M}\) solution of hypobromite ion $\left(\mathrm{BrO}^{-} ; K_{b}=4.0 \times 10^{-6}\right) .\( What is the \)\mathrm{pH}$ of this solution?

Short Answer

Expert verified
The molar concentration of \(\mathrm{OH}^{-}\) in the \(0.724\ \mathrm{M}\) solution of hypobromite ion \(\mathrm{BrO}^{-}\) is approximately \(1.702 \times 10^{-3}\ \mathrm{M}\). The \(\mathrm{pH}\) of this solution is approximately 11.23.

Step by step solution

01

Write the equilibrium expression for the reaction of \(\mathrm{BrO}^{-}\) with water

The reaction of the hypobromite ion (\(\mathrm{BrO}^{-}\)) with water can be written as: \[\mathrm{BrO}^{-} + \mathrm{H_{2}O} \rightleftharpoons \mathrm{OH}^{-} + \mathrm{HBrO}\] Now, we can write the expression for the \(K_{b}\) constant: \[K_{b} = \frac{[\mathrm{OH}^{-}][\mathrm{HBrO}]}{[\mathrm{BrO}^{-}]}\]
02

Define the initial and equilibrium concentrations

Let's assume that the change in the concentration of \(\mathrm{BrO}^{-}\) is represented by \(x\). At equilibrium, the concentrations will be: \([\mathrm{BrO}^{-}] = 0.724 - x\) \([\mathrm{OH}^{-}] = x\) \([\mathrm{HBrO}] = x\)
03

Substitute the equilibrium concentrations into the \(K_{b}\) expression

Plugging in the equilibrium concentrations into the \(K_{b}\) expression, we get: \[4.0 \times 10^{-6} = \frac{x\cdot x}{0.724 - x}\]
04

Solve for x

Since \(K_{b}\) is very small, we can make the assumption that \(x << 0.724\). In that case, we can approximate the denominator: \[0.724 - x \approx 0.724\] Now, solving for \(x\), we have: \[x^2 = 4.0 \times 10^{-6} \times 0.724\] \[x^2 = 2.896 \times 10^{-6}\] \[x \approx 1.702 \times 10^{-3}\] As \(x\) represents the concentration of \(\mathrm{OH}^{-}\), we have: \([\mathrm{OH}^{-}] \approx 1.702 \times 10^{-3}\ \mathrm{M}\)
05

Calculate the \(\mathrm{pOH}\) and then the \(\mathrm{pH}\)

Now that we have the concentration of \(\mathrm{OH}^{-}\), we can calculate the \(\mathrm{pOH}\) as: \[\mathrm{pOH} = -\log{[\mathrm{OH}^{-}]}\] \[\mathrm{pOH} = -\log{(1.702 \times 10^{-3})} \approx 2.77\] And finally, we can determine the \(\mathrm{pH}\) as: \[\mathrm{pH} = 14 - \mathrm{pOH} = 14 - 2.77 \approx 11.23\] Therefore, the \(\mathrm{pH}\) of the given solution is approximately 11.23.

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Most popular questions from this chapter

\(\mathrm{NH}_{3}(g)\) and \(\mathrm{HCl}(g)\) react to form the ionic solid \(\mathrm{NH}_{4} \mathrm{Cl}(s) .\) Which substance is the Brønsted-Lowry acid in this reaction? Which is the Brønsted-Lowry base?

Using data from Appendix \(D\), calculate \(p O H\) and \(p H\) for each (a) \(0.080 M\) potassium hypobromite of the following solutions: \((\mathrm{KBrO}),\) (b) \(0.150 \mathrm{M}\) potassium hydrosulfide \((\mathrm{KHS}),(\mathbf{c})\) a mixture that is \(0.25 \mathrm{M}\) in potassium nitrite \(\left(\mathrm{KNO}_{2}\right)\) and \(0.15 \mathrm{M}\) in magnesium nitrite \(\left(\mathrm{Mg}\left(\mathrm{NO}_{2}\right)_{2}\right)\).

Addition of phenolphthalein to an unknown colorless solution does not cause a color change. The addition of bromthymol blue to the same solution leads to a yellow color. (a) Is the solution acidic, neutral, or basic? (b) Which of the following can you establish about the solution: (i) A minimum \(\mathrm{pH}\), (ii) A maximum \(\mathrm{pH}\), or (iii) A specific range of pH values? (c) What other indicator or indicators would you want to use to determine the \(\mathrm{pH}\) of the solution more precisely?

(a) Which of the following is the stronger Brønsted-Lowry acid, \(\mathrm{HClO}_{3}\) or \(\mathrm{HClO}_{2} ?\) (b) Which is the stronger Brønsted-

Identify the Brønsted-Lowry acid and the Brønsted-Lowry base on the left side of each of the following equations, and also identify the conjugate acid and conjugate base of each on the right side: (a) $\mathrm{NH}_{4}^{+}(a q)+\mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{HCN}(a q)+\mathrm{NH}_{3}(a q)$ (b) $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons$ $$ \left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ (c) \(\mathrm{HCOOH}(a q)+\mathrm{PO}_{4}^{3-}(a q) \rightleftharpoons\) $$ \mathrm{HCOO}^{-}(a q)+\mathrm{HPO}_{4}^{2-}(a q) $$
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