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Chapter 17: Problem 87

Equal quantities of \(0.010 \mathrm{M}\) solutions of an acid HA and a base \(\mathrm{B}\) are mixed. The \(\mathrm{pH}\) of the resulting solution is 9.2 . (a) Write the chemical equation and equilibrium-constant expression for the reaction between HA and 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 chemical equation for the reaction between the acid HA and the base B is \(HA + B^- \rightleftharpoons A^- + HB\), with the equilibrium constant expression \(K = \frac{[A^-][HB]}{[HA][B^-]}\). The equilibrium constant K for the reaction between HA and B is 12.65. The value of Kb for B is 1.25 x 10^(-10).

Step by step solution

01

Write the chemical equation and equilibrium constant expression

First, let's write the chemical equation for the reaction between the acid HA and the base B. Since we are dealing with an acid-base reaction, we can write the equation as follows: \(HA + B^- \rightleftharpoons A^- + HB\) Now, let's write the equilibrium constant expression for this reaction: \(K = \frac{[A^-][HB]}{[HA][B^-]}\)
02

Calculate the concentration of H+ ions

The given pH of the resulting solution is 9.2. We can use the pH to calculate the concentration of H+ ions using the formula: \([H^+] = 10^{-pH}\) \([H^+] = 10^{-9.2} = 6.31 \times 10^{-10} M\)
03

Calculate the concentration of A- ions

Now, let's use the given Ka value for HA to find the concentration of A- ions. The equation for Ka is: \(K_a = \frac{[A^-][H^+]}{[HA]}\) Given the Ka value as 8.0 x 10^(-5), we can rearrange the equation and solve for the concentration of A- ions: \([A^-] = \frac{K_a[HA]}{[H^+]}\) \([A^-] = \frac{(8.0 \times 10^{-5})(0.010)}{6.31 \times 10^{-10}}\) \([A^-] = 1.27 \times10^{-4} M\)
04

Calculate the equilibrium constant for the reaction between HA and B

Now, let's find the equilibrium constant for the reaction between HA and B using the equilibrium constant expression and the concentrations of the reacting species: \(K = \frac{[A^-][HB]}{[HA][B^-]}\) Given that the reaction was between equal quantities of HA and B, their concentrations will be equal at equilibrium: \([HB] = [B^-] = [HA] - [A^-]\) Using the known values, \(K = \frac{[A^-][HA] - [A^-]}{[HA]^2 - [A^-][HA]}\) \(K = \frac{(1.27 \times 10^{-4})(0.010 - 1.27 \times 10^{-4})}{(0.010)^2 - (1.27 \times 10^{-4})(0.010)}\) \(K = 12.65\)
05

Calculate the value of Kb for B

Lastly, we will find the value of Kb for B. We can use the relationship between Ka and Kb for conjugate acid-base pairs, which is: \(K_a \times K_b = K_w\) Where Kw is the ion product of water (1.0 x 10^(-14) at 25°C). Rearranging the equation, we can solve for Kb: \(K_b = \frac{K_w}{K_a}\) \(K_b = \frac{1.0 \times 10^{-14}}{8.0 \times 10^{-5}}\) \(K_b = 1.25 \times 10^{-10}\) So, the value of Kb for B is 1.25 x 10^(-10).

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Most popular questions from this chapter

For each of the following slightly soluble salts, write the net ionic equation, if any, for reaction with a strong acid: (a) MnS, (b) \(\mathrm{PbF}_{2}\), (c) \(\mathrm{AuCl}_{3}\) (d) \(\mathrm{Hg}_{2} \mathrm{C}_{2} \mathrm{O}_{4},\) (e) CuBr.

(a) A 0.1044-g sample of an unknown monoprotic acid requires $22.10 \mathrm{~mL}\( of \)0.0500 \mathrm{M} \mathrm{NaOH}$ to reach the end point. What is the molar mass of the unknown? (b) As the acid is titrated, the \(\mathrm{pH}\) of the solution after the addition of \(11.05 \mathrm{~mL}\) of the base is \(4.89 .\) What is the \(K_{a}\) for the acid? (c) Using Appendix D, suggest the identity of the acid.

Compare the titration of a strong, monoprotic acid with a strong base to the titration of a weak, monoprotic acid with a strong base. Assume the strong and weak acid solutions initially have the same concentrations. Indicate whether the following statements are true or false. (a) More base is required to reach the equivalence point for the strong acid than the weak acid. \((\mathbf{b})\) The \(\mathrm{pH}\) at the beginning of the titration is lower for the weak acid than the strong acid. \((\mathbf{c})\) The \(\mathrm{pH}\) at the equivalence point is 7 no matter which acid is titrated.

Furoic acid \(\left(\mathrm{HC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}\right)\) has a \(K_{a}\) value of \(6.76 \times 10^{-4}\) at \(25^{\circ} \mathrm{C}\). Calculate the \(\mathrm{pH}\) at \(25^{\circ} \mathrm{C}\) of \((\mathbf{a})\) a solution formed by adding \(30.0 \mathrm{~g}\) of furoic acid and \(25.0 \mathrm{~g}\) of sodium furoate \(\left(\mathrm{NaC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}\right)\) to enough water to form \(0.300 \mathrm{~L}\) of solution, (b) a solution formed by mixing \(20.0 \mathrm{~mL}\) of \(0.200 \mathrm{M}\) $\mathrm{HC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}\( and \)30.0 \mathrm{~mL}\( of \)0.250 \mathrm{M} \mathrm{NaC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}$ and diluting the total volume to \(125 \mathrm{~mL},(\mathbf{c})\) a solution prepared by adding $25.0 \mathrm{~mL}\( of \)1.00 \mathrm{M} \mathrm{NaOH}\( solution to \)100.0 \mathrm{~mL}\( of \)0.100 \mathrm{M} \mathrm{HC}_{5} \mathrm{H}_{3} \mathrm{O}_{3}$

A 10.0-mL sample of \(0.250 \mathrm{MHNO}_{3}\) solution is titrated with $0.100 M$ KOH solution. Calculate the pH of the solution after the following volumes of base have been added: (a) \(20.0 \mathrm{~mL}\), (b) \(24.9 \mathrm{~mL}\) (c) \(25.0 \mathrm{~mL}\) (d) \(25.1 \mathrm{~mL}\), (e) \(30.0 \mathrm{~mL}\).
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