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Chapter 14: Problem 151

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

Short Answer

Expert verified
The solubility of benzoic acid in water at 25°C is calculated using the given Ka value and pH, resulting in a solubility of \(6.85 \times 10^{-3}\) mol/L.

Step by step solution

01

Write the Ka expression for the dissociation of benzoic acid

Benzoic acid (\(HA\)) dissociates in water as follows: \(HA \rightleftharpoons H^{+} + A^{-}\) Given the acid dissociation constant (Ka), we can write the Ka expression as: \(K_{a} = \frac{[H^{+}][A^{-}]}{[HA]}\)
02

Relate the ion concentrations in the solution

Let's assume the initial concentration of benzoic acid, [HA], is S mol/L and it dissociates to form H⁺ and A⁻ ions. So, the change in the concentrations of all species in the solution will be: [H⁺] = [A⁻] = x [HA] = S - x
03

Calculate H⁺ concentration using the given pH

We are given that the pH of the solution is 2.80. We can calculate the H⁺ ion concentration by using the pH formula as: \(pH = -\log[H^{+}]\) Now substituting the value of pH, we get: 2.80 = -\log[H⁺] Calculate the [H⁺]: [H⁺] = 10^(-2.80)
04

Substitute the ion concentrations into the Ka expression

Now we can substitute the ion concentrations into the Ka expression: \(K_{a} =\frac{[H^{+}][A^{-}]}{[HA]} = \frac{x (x)}{(S-x)}\)
05

Use the given Ka value and solve for x and S

Given that Ka = 6.4 × 10⁻⁵, we can substitute this value and the calculated [H⁺] into the equation and solve for S: \(6.4 \times 10^{-5} = \frac{x^2}{S-x}\) As the dissociation of the benzoic acid is small compared to the solubility, we can assume that x is small compared to S and thus simplify the expression: \(6.4 \times 10^{-5} \approx \frac{x^2}{S}\) Now rearrange the equation to calculate S: \(S = \frac{x^2}{6.4 \times 10^{-5}}\)
06

Calculate the solubility of benzoic acid

Now that we have the expression to find the solubility S, we'll substitute the value of x calculated in step 3: S = \(\frac{(10^{-2.80})^2}{6.4 \times 10^{-5}}\) Finally, we can calculate the value of S: S = 6.85 × 10⁻³ mol/L The solubility of benzoic acid in water at 25°C is 6.85 × 10⁻³ mol/L.

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Most popular questions from this chapter

Arrange the following 0.10\(M\) solutions in order from most acidic to most basic. See Appendix 5 for \(K_{\mathrm{a}}\) and \(K_{\mathrm{b}}\) values. $$ \mathrm{CaBr}_{2}, \mathrm{KNO}_{2}, \mathrm{HClO}_{4}, \quad \mathrm{HNO}_{2}, \quad \mathrm{HONH}_{3} \mathrm{ClO}_{4} $$

Write the reaction and the corresponding \(K_{\mathrm{b}}\) equilibrium expression for each of the following substances acting as bases in water. a. \(\mathrm{NH}_{3}\) b. \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\)

The 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{mM},\( whereas his \)\mathrm{CO}_{2}\( levels were 1.63 \)\mathrm{mM}$ . 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?

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Calculate the \(\mathrm{pH}\) of a $0.050-M\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}\( solution \)\left(K_{\mathrm{b}}=\right.$ \(1.3 \times 10^{-3} )\)
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