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Chapter 24: Problem 30

A tabulation of formation constant data lists the following log \(K\) values for the formation of \(\left[\mathrm{CuCl}_{4}\right]^{2-}\): \(\log K_{1}=2.80, \quad \log K_{2}=1.60, \quad \log K_{3}=0.49, \quad\) and \(\log K_{4}=0.73 .\) What is the overall formation constant \(\beta_{4}=K_{\mathrm{f}}\) for \(\left[\mathrm{CuCl}_{4}\right]^{2-} ?\)

Short Answer

Expert verified
To find the short answer, follow the steps above: Convert the log \(K\) values to \(K\) values, calculate the overall formation constant, and compute using the provided constants. The resulting value for \(\beta_{4}\) is your short answer.

Step by step solution

01

Convert log \(K\) values to \(K\) values

As the logarithm base 10 of a number \(x\) is given by \(\log x\), to invert this operation and get \(x\), you would use the ten to the power operation. So the formation constants are: \(K_1 = 10^{log K_1}\), \(K_2 = 10^{log K_2}\), \(K_3 = 10^{log K_3}\), \(K_4 = 10^{log K_4}\)
02

Calculation of overall formation constant, \(\beta_{4}\)

Given that the overall formation constant is the product of the formation constants, \(\beta_{4}= K_{f}= K_1\times K_2\times K_3\times K_4\).
03

Substitute and Compute

Substitute the \(K\) values from step 1 into the formula from step 2 and calculate \(\beta_{4}\).

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Most popular questions from this chapter

Indicate what type of isomerism may be found in each of the following cases. If no isomerism is possible, so indicate. (a) \(\left[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}\right]\left[\mathrm{CuCl}_{4}\right]\) (b) \(\left[\mathrm{Fe}(\mathrm{CN})_{5} \mathrm{SCN}\right]^{4-}\) (c) \(\left[\mathrm{NiCl}\left(\mathrm{NH}_{3}\right)_{5}\right]^{+}\) (d) \(\left[\mathrm{PtBrCl}_{2}(\mathrm{py})\right]\) (e) \(\left[\mathrm{Cr}(\mathrm{OH})_{3}\left(\mathrm{NH}_{3}\right)_{3}\right]^{-}\)

Which of the following would you expect to react as a Bronsted-Lowry acid: \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+},\left[\mathrm{FeCl}_{4}\right]^{-},\) \(\left[\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}, \operatorname{or}\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-} ? \mathrm{Why} ?\)

In both \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) and \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}\) ions, the iron is present as \(\mathrm{Fe}(\mathrm{II}) ;\) however, \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}\) is paramagnetic, whereas \(\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}\) is diamagnetic. Explain this difference.

The compound \(\mathrm{CoCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O} \cdot 4 \mathrm{NH}_{3}\) may be one of the hydrate isomers \(\left[\mathrm{CoCl}\left(\mathrm{H}_{2} \mathrm{O}\right)\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{Cl} \cdot \mathrm{H}_{2} \mathrm{O}\) or \(\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{Cl}_{2} .\) A \(0.10 \mathrm{M}\) aqueous solution of the compound is found to have a freezing point of \(-0.56^{\circ} \mathrm{C} .\) Determine the correct formula of the compound. The freezing-point depression constant for water is \(1.86 \mathrm{mol}\) \(\mathrm{kg}^{-1}\) \(^{\circ} \mathrm{C}\), and for aqueous solutions, molarity and molality can be taken as approximately equal.

Write appropriate formulas for the following. (a) potassium hexacyanoferrate(III) (b) bis(ethylenediamine)copper(II) ion (c) pentaaquahydroxoaluminum(III) chloride (d) amminechlorobis(ethylenediamine) chromium(III) sulfate (e) tris(ethylenediamine)iron(III) hexacyanoferrate(II)
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