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Chapter 14: Problem 68

A solution is prepared by dissolving \(0.56 \mathrm{~g}\) benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}, K_{\mathrm{a}}=6.4 \times 10^{-5}\right)\) in enough water to make \(1.0 \mathrm{~L}\) of solution. Calculate \(\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\right],\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}\right],\left[\mathrm{H}^{+}\right]\), \(\left[\mathrm{OH}^{-}\right]\), and the \(\mathrm{pH}\) of this solution.

Short Answer

Expert verified
The molar concentrations of the species in the solution are: \([\text{C}_{6}\text{H}_{5}\text{CO}_{2}\text{H}] = 4.33 \times 10^{-3} \, M\) \([\text{C}_{6}\text{H}_{5}\text{CO}_{2}^{-}] = 2.65 \times 10^{-4} \, M\) \([\text{H}^{+}] = 2.65 \times 10^{-4} \, M\) \([\text{OH}^{-}] = 3.77 \times 10^{-11} \, M\) \(\text{pH} = 3.58\)

Step by step solution

01

Calculate the initial concentration of benzoic acid

Begin by calculating the initial molar concentration of benzoic acid in the solution using the given mass (0.56 g) and volume (1.0 L). The molar mass of benzoic acid is \(C_{6}H_{5}CO_{2}H = 12.01(6)+1.01(5)+12.01+16.00(2)+1.01 = 122.12 \, g/mol\). Using the mass (\(m = 0.56 \, g\)) and molar mass (\(M = 122.12 \, g/mol\)), we can determine the initial moles of benzoic acid: \(\text{moles} = \dfrac{m}{M} = \dfrac{0.56 \, g}{122.12 \, g/mol} = 4.59 \times 10^{-3} \, mol\) Now, divide the moles by the volume of the solution (1.0 L) to get the initial concentration: \([\text{C}_{6}\text{H}_{5}\text{CO}_{2}\text{H}]_i = \dfrac{4.59 \times 10^{-3} \, mol}{1.0 \, L} = 4.59 \times 10^{-3} \, M\)
02

Write the Ka expression

Write the Ka expression for the acid dissociation reaction: \(\text{Ka} = \dfrac{[\text{C}_{6}\text{H}_{5}\text{CO}_{2}^{-}][\text{H}^{+}]}{[\text{C}_{6}\text{H}_{5}\text{CO}_{2}\text{H}]}\)
03

Apply the Ka expression to the system

Assume that x moles of benzoic acid dissociate in the solution. Therefore, we can express the concentrations at equilibrium in terms of x using the initial concentration of benzoic acid. \([\text{C}_{6}\text{H}_{5}\text{CO}_{2}\text{H}] = (4.59 \times 10^{-3} - x) \, M\) \([\text{C}_{6}\text{H}_{5}\text{CO}_{2}^{-}] = x \, M\) \([\text{H}^{+}] = x \, M\) Now, substitute these expressions into the Ka equation: \(\text{Ka} = \dfrac{x^2}{(4.59 \times 10^{-3} - x)}\)
04

Solve for x

Given the value of Ka is \(6.4 \times 10^{-5}\), we can solve for x: \(6.4 \times 10^{-5} = \dfrac{x^2}{(4.59 \times 10^{-3} - x)}\) Since the value of Ka is quite small, we can make the approximation that x is much smaller than \(4.59 \times 10^{-3}\) and simplify the above equation as: \(6.4 \times 10^{-5} \approx \dfrac{x^2}{4.59 \times 10^{-3}}\) Solve for x: \(x \approx \sqrt{6.4 \times 10^{-5} \times 4.59 \times 10^{-3}} = 2.65 \times 10^{-4}\)
05

Calculate the concentrations

Use the value of x to calculate the concentrations at equilibrium: \([\text{C}_{6}\text{H}_{5}\text{CO}_{2}\text{H}] \approx 4.59 \times 10^{-3} - 2.65 \times 10^{-4} = 4.33 \times 10^{-3} \, M\) \([\text{C}_{6}\text{H}_{5}\text{CO}_{2}^{-}] \approx 2.65 \times 10^{-4} \, M\) \([\text{H}^{+}] \approx 2.65 \times 10^{-4} \, M\)
06

Calculate the OH⁻ concentration and pH

With the H⁺ concentration, we can use the ion-product constant for water (Kw = \(1.0 \times 10^{-14}\)) to calculate the OH⁻ concentration: \([\text{OH}^{-}] = \dfrac{\text{Kw}}{[\text{H}^{+}]} = \dfrac{1.0 \times 10^{-14}}{2.65 \times 10^{-4}} = 3.77 \times 10^{-11} \, M\) Using the H⁺ concentration, we can calculate the pH of the solution using the formula pH = \(-log([\text{H}^{+}])\): \(\text{pH} = -log(2.65 \times 10^{-4}) = 3.58\)
07

Final results

The molar concentrations of the species in the solution are as follows: \([\text{C}_{6}\text{H}_{5}\text{CO}_{2}\text{H}] = 4.33 \times 10^{-3} \, M\) \([\text{C}_{6}\text{H}_{5}\text{CO}_{2}^{-}] = 2.65 \times 10^{-4} \, M\) \([\text{H}^{+}] = 2.65 \times 10^{-4} \, M\) \([\text{OH}^{-}] = 3.77 \times 10^{-11} \, M\) \(\text{pH} = 3.58\)

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Most popular questions from this chapter

An unknown salt is either \(\mathrm{NaCN}, \mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\), NaF, \(\mathrm{NaCl}\), or \(\mathrm{NaOCl}\). When \(0.100\) mole of the salt is dissolved in \(1.00 \mathrm{~L}\) of solution, the \(\mathrm{pH}\) of the solution is \(8.07\). What is the identity of the salt?

What are the major species present in the following mixtures of bases? a. \(0.050 \mathrm{M} \mathrm{NaOH}\) and \(0.050 \mathrm{M} \mathrm{LiOH}\) b. \(0.0010 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\) and \(0.020 \mathrm{M} \mathrm{RbOH}\) What is \(\left[\mathrm{OH}^{-}\right]\) and the \(\mathrm{pH}\) of each of these solutions?

Calculate the \(\mathrm{pH}\) of a \(0.200-M\) solution of \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHF}\). Hint: \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHF}\) is a salt composed of \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\) and \(\mathrm{F}^{-}\) ions. The principal equilibrium in this solution is the best acid reacting with the best base; the reaction for the principal equilibrium is \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons\) \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}(a q)+\mathrm{HF}(a q) \quad K=8.2 \times 10^{-3}\)

A \(0.15-M\) solution of a weak acid is \(3.0 \%\) dissociated. Calculate \(K_{\mathrm{a}}\)

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. \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\)
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