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Chapter 23: Problem 62

Equation \((23.18),\) which represents the chromatedichromate equilibrium, is actually the sum of two equilibrium expressions. The first is an acid-base reaction, \(\mathrm{H}^{+}+\mathrm{CrO}_{4}^{2-} \rightleftharpoons \mathrm{HCrO}_{4}^{-}\). The second reaction involves elimination of a water molecule between two \(\mathrm{HCrO}_{4}^{-}\) ions (a dehydration reaction), \(2 \mathrm{HCrO}_{4}^{-} \rightleftharpoons \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{H}_{2} \mathrm{O} .\) If the ionization constant, \(K_{\mathrm{a}},\) for \(\mathrm{HCrO}_{4}^{-}\) is \(3.2 \times 10^{-7},\) what is the value of \(K\) for the dehydration reaction?

Short Answer

Expert verified
The equilibrium constant, \( K \), for the dehydration reaction is approximately \( \frac{1}{3.2 \times 10^{-7}} \)

Step by step solution

01

Identify relevant equations

This problem involves two reactions-\n\nReaction 1: \(\mathrm{H}^{+}+\mathrm{CrO}_{4}^{2-} \rightleftharpoons \mathrm{HCrO}_{4}^{-}\) with a known ionization constant, \( K_a = 3.2 \times 10^{-7}\)\n\nReaction 2: \(2 \mathrm{HCrO}_{4}^{-} \rightleftharpoons \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}+\mathrm{H}_{2}\mathrm{O}\), for which we need to find the equilibrium constant, \( K \).\n\nThe sum of these reactions is the complete chromatedichromate equilibrium reaction, defined in Equation (23.18).
02

Use properties of equilibrium constants

The equilibrium constant for the net reaction obtained by adding multiple reactions is the product of the equilibrium constants for the individual reactions. Suppose \( K' \) is the equilibrium constant for the overall reaction, and \( K \) and \( K_a \) are equilibrium constants for reactions 1 and 2, respectively. Then, by the property of equilibrium constants, we have: \[ K' = K \cdot K_a \]
03

Use the given ionisation constant

Substitute the ionization constant \( K_a \) of the first reaction into the formula: \[ K' = K \cdot 3.2 \times 10^{-7} \]
04

Calculate K

As we know the equilibrium constant for the complete reaction can only be 1, we can substitute this into the formula and solve for \( K \). \[ 1 = K \cdot 3.2 \times 10^{-7} \] Solving for \( K \), we get \( K = \frac{1}{3.2 \times 10^{-7}} \)
05

Final Answer

Calculate the value using a calculator to get the final result.

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Most popular questions from this chapter

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