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Chapter 14: Problem 149

At \(25^{\circ} \mathrm{C}\), a saturated solution of benzoic acid \(\left(K_{\mathrm{a}}=6.4 \times\right.\) \(10^{-5}\) ) has a pH of \(2.80 .\) Calculate the water solubility of benzoic acid in moles per liter.

Short Answer

Expert verified
The water solubility of benzoic acid at \(25^{\circ} \mathrm{C}\) is approximately \(6.4 \times 10^{-5}\) moles per liter.

Step by step solution

01

Write the reaction for benzoic acid and its dissociation

Benzoic acid (HC7H5O2) will dissociate in water as follows: \[HC7H5O2_{(aq)} \rightleftharpoons H^+_{(aq)} + C7H5O2^{-}_{(aq)}\]
02

Write the expression for the Ka value of benzoic acid

Using the reaction from Step 1, the Ka expression for benzoic acid is: \[K_{a} = \frac{[H^+][C7H5O2^{-}]}{[HC7H5O2]}\]
03

Find the concentration of \(H^+\) ions from the pH value

The pH is given as \(2.80\). We can find the concentration of \(H^+\) ions using the pH formula: \[pH = -\log_{10} [H^+]\] Rearrange the formula and solve for \([H^+]\): \[[H^+] = 10^{(-pH)}\] \[[H^+] = 10^{(-2.80)}\] \[[H^+] = 1.58 \times 10^{-3} \mathrm{M}\]
04

Express equilibrium concentrations in terms of solubility

Let the solubility of benzoic acid in water be represented by variable 's' in moles per liter. At equilibrium, the concentrations are as follows: \[ [HC7H5O2] = s \mathrm{M} \] \[ [H^+] = (1.58 \times 10^{-3} + s) \mathrm{M} \] \[ [C7H5O2^{-}] = s \mathrm{M} \]
05

Substitute equilibrium concentrations into Ka expression and solve for solubility

Plug the equilibrium concentrations from Step 4 into the Ka expression from Step 2: \[6.4 \times 10^{-5} = \frac{(1.58 \times 10^{-3} + s)(s)}{s}\] Since the solubility of benzoic acid (s) is significantly smaller than \([H^+]\), we can approximate by considering s to be negligible compared to \([H^+]\): \[6.4 \times 10^{-5} \approx \frac{(1.58 \times 10^{-3})(s)}{s}\] Now, solve for s: \[s \approx 6.4 \times 10^{-5} \mathrm{M}\] So, the water solubility of benzoic acid at \(25^{\circ} \mathrm{C}\) is approximately \(6.4 \times 10^{-5}\) moles per liter.

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Most popular questions from this chapter

Sodium azide \(\left(\mathrm{NaN}_{3}\right)\) is sometimes added to water to kill bacteria. Calculate the concentration of all species in a \(0.010-M\) solution of \(\mathrm{NaN}_{3}\). The \(K_{\mathrm{a}}\) value for hydrazoic acid \(\left(\mathrm{HN}_{3}\right)\) is \(1.9 \times 10^{-5}\)

The \(\mathrm{pH}\) of human blood is steady at a value of approximately \(7.4\) owing to the following equilibrium reactions: \(\mathrm{CO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightleftharpoons \mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}^{+}(a q)\) Acids formed during normal cellular respiration react with the \(\mathrm{HCO}_{3}^{-}\) to form carbonic acid, which is in equilibrium with \(\mathrm{CO}_{2}(a q)\) and \(\mathrm{H}_{2} \mathrm{O}(l) .\) During vigorous exercise, a person's \(\mathrm{H}_{2} \mathrm{CO}_{3}\) blood levels were \(26.3 \mathrm{~m} M\), whereas his \(\mathrm{CO}_{2}\) levels were \(1.63 \mathrm{~m} M .\) On resting, the \(\mathrm{H}_{2} \mathrm{CO}_{3}\) levels declined to \(24.9 \mathrm{~m} M\). What was the \(\mathrm{CO}_{2}\) blood level at rest?

Given that the \(K_{\mathrm{a}}\) value for acetic acid is \(1.8 \times 10^{-5}\) and the \(K_{\text {a }}\) value for hypochlorous acid is \(3.5 \times 10^{-8}\), which is the stronger base, \(\mathrm{OCl}^{-}\) or \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}\) ?

Will the following oxides give acidic, basic, or neutral solutions when dissolved in water? Write reactions to justify your answers. a. \(\mathrm{CaO}\) b. \(\mathrm{SO}_{2}\) c. \(\mathrm{Cl}_{2} \mathrm{O}\)

Will the following oxides give acidic, basic, or neutral solutions when dissolved in water? Write reactions to justify your answers. a. \(\mathrm{Li}_{2} \mathrm{O}\) b. \(\mathrm{CO}_{2}\) c. \(\mathrm{SrO}\)
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