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Chapter 18: Problem 8

In the qualitative cation analysis procedure, \(\mathrm{Bi}^{3+}\) is detected by the appearance of a white precipitate of bismuthyl hydroxide, \(\mathrm{BiOOH}(\mathrm{s})\): \(\mathrm{BiOOH}(\mathrm{s}) \rightleftharpoons \mathrm{BiO}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})\) \(K_{\mathrm{sp}}=4 \times 10^{-10}\) Calculate the \(\mathrm{pH}\) of a saturated aqueous solution of \(\mathrm{BiOOH}\).

Short Answer

Expert verified
The pH of a saturated aqueous solution of BiOOH is 9.30.

Step by step solution

01

Write the equilibrium expression

The equilibrium reaction can be written as: \(\mathrm{BiOOH}(\mathrm{s}) \rightleftharpoons\mathrm{BiO}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})\). The solubility product expression becomes: \[ K_{sp} = [\mathrm{BiO}^+][\mathrm{OH}^-] \]
02

Set up the solubility relationship

Let's denote the solubility of BiOOH as \(s\). Then, \([\mathrm{BiO}^+]=s\) and \([\mathrm{OH}^-]=s\). Substituting these into the \(K_{sp}\) expression we get: \( K_{sp}=s^2 = 4 \times 10^{-10}\). Solving for \(s\) we get \(s=\sqrt{4 \times 10^{-10}} = 2 \times 10^{-5}\) M
03

Calculating pH

Now, since \(\mathrm{pOH} = -\log[\mathrm{OH}^-]\), we can substitute \(s = [\mathrm{OH}^-]\) into the formula to calculate the \(\mathrm{pOH} = -\log(2 \times 10^{-5}) \approx 4.70\). Then, because \(\mathrm{pOH} + \mathrm{pH} = 14\) in water at 25 C, we can substitute the \(\mathrm{pOH}\) into the equation to get \(\mathrm{pH} = 14 - \mathrm{pOH} = 14 - 4.70 = 9.30\).
04

Rounding to correct significant figures

Following the significant figure rules, the final answer can be rounded to two decimal places, giving a \(\mathrm{pH} = 9.30\).

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