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Chapter 17: Problem 85

Equal quantities of \(0.010 \mathrm{M}\) solutions of an acid \(\mathrm{HA}\) and a base \(\mathrm{B}\) are mixed. The \(\mathrm{pH}\) of the resulting solution is \(9.2 .(\mathrm{a})\) Write the equilibrium equation and equilibrium-constant expression for the reaction between \(\mathrm{HA}\) and \(\mathrm{B}\). (b) If \(K_{a}\) for HA is \(8.0 \times 10^{-5}\), what is the value of the equilibrium constant for the reaction between HA and \(B\) ? (c) What is the value of \(K_{b}\) for B?

Short Answer

Expert verified
The equilibrium constant for the reaction between HA and B is approximately \(1.56 \times 10^{6}\), and the Kb value for base B is approximately \(1.25 \times 10^{-10}\).

Step by step solution

01

Write the reaction and equilibrium equation

The reaction between acid HA and base B can be written as follows: \[ \mathrm{HA} + \mathrm{B}^{-} \rightleftharpoons \mathrm{A}^{-} + \mathrm{HB} \] The equilibrium constant expression for this reaction can be written as: \[ K = \frac{ [\mathrm{A}^{-}][\mathrm{HB}]} { [\mathrm{HA}][\mathrm{B}^{-}]} \]
02

Note the given values

We have the following given values: - The pH of the resulting solution is 9.2 - The Ka value for HA is \( 8.0 \times 10^{-5} \)
03

Calculate the concentration of \(\mathrm{A}^{-}\) and \(\mathrm{HB}\) in the resulting solution

We know that pH is related to the concentration of hydrogen ions as: \[ \mathrm{pH} = - \log[H^{+}] \] We are given that the pH of the resulting solution is 9.2. Hence, we can calculate the concentration of hydrogen ions: \[ H^{+} = 10^{-\mathrm{pH}} = 10^{-9.2} \] Since the resulting solution has negligible hydrogen ions, we can assume that most of the acid HA has dissociated into A- and HB. So, the concentrations of A- and HB in the resulting solution will be approximately equal to the initial concentration of HA: \[ [\mathrm{A}^{-}] \approx [\mathrm{HB}] \approx 0.010 \mathrm{M} \]
04

Calculate the equilibrium constant for the reaction between HA and B

We can now use the equilibrium constant expression: \[ K = \frac{ [\mathrm{A}^{-}][\mathrm{HB}]} { [\mathrm{HA}][\mathrm{B}^{-}]} \] We know that: \[ [\mathrm{A}^{-}] \approx [\mathrm{HB}] \approx 0.010 \mathrm{M} \] As equal quantities of HA and B are mixed, their initial concentrations can be considered equal: \[ [\mathrm{HA}] \approx [\mathrm{B}^{-}] \] So, the expression for K can be simplified to: \[ K = \frac{ (0.010)^{2}} { [\mathrm{HA}]^{2}} \] Substituting Ka for HA, we have: \[ K = \frac{(0.010)^{2}}{(8.0 \times 10^{-5})^{2}} \] \[ K \approx 1.56 \times 10^{6} \] So, the equilibrium constant for the reaction between HA and B is approximately \(1.56 \times 10^{6}\).
05

Calculate the Kb value for B

We know that for a conjugate acid-base pair, the following relationship holds true: \[ K_{a} \cdot K_{b} = K_{w} \] where \(K_{w}\) is the ion product of water, which is \(1.0 \times 10^{-14}\) at 25°C. We are given the Ka value for HA and we just found the equilibrium constant for the reaction between HA and B. Now, we need to find Kb for B. Re-write the above equation as: \[ K_{b} = \frac{K_{w}}{ K_{a} } \] Substitute the given values and the calculated constant: \[ K_{b} = \frac{(1.0 \times 10^{-14})}{(8.0 \times 10^{-5})} \] \[ K_{b} \approx 1.25 \times 10^{-10} \] So, the Kb value for base B is approximately \(1.25 \times 10^{-10}\).

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Most popular questions from this chapter

Suppose you want to do a physiological experiment that calls for a pH 6.50 buffer. You find that the organism with which you are working is not sensitive to the weak acid \(\mathrm{H}_{2} \mathrm{X}\) \(\left(K_{a 1}=2 \times 10^{-2} ; K_{a 2}=5.0 \times 10^{-7}\right)\) or its sodium salts. You have available a \(1.0 \mathrm{M}\) solution of this acid and a \(1.0 \mathrm{M}\) solution of \(\mathrm{NaOH}\). How much of the \(\mathrm{NaOH}\) solution should be added to \(1.0 \mathrm{~L}\) of the acid to give a buffer at \(\mathrm{pH} 6.50 ?\) (Ignore any volume change.)

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From the value of \(K_{f}\) listed in Table \(17.1,\) calculate the concentration of \(\mathrm{Ni}^{2+}\) in \(1.0 \mathrm{~L}\) of a solution that contains a total of \(1 \times 10^{-3} \mathrm{~mol}\) of nickel(II) ion and that is \(0.20 \mathrm{M}\) in \(\mathrm{NH}_{3}\)
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