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Chapter 14: Problem 65

For propanoic acid \(\left(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}, K_{\mathrm{a}}=1.3 \times 10^{-5}\right)\), determine the concentration of all species present, the \(\mathrm{pH}\), and the percent dissociation of a \(0.100 M\) solution.

Short Answer

Expert verified
At equilibrium, the concentrations of the species in a 0.100 M propanoic acid solution are: [HC₃H₅O₂] ≈ 0.0964 M, [H⁺] ≈ 0.0036 M, and [C₃H₅O₂⁻] ≈ 0.0036 M. The pH of this solution is approximately 2.44, and the percent dissociation of propanoic acid is ≈ 3.6%.

Step by step solution

01

1. Write the dissociation equation

The dissociation equation for propanoic acid is: HC₃H₅O₂(aq) ⇌ H⁺(aq) + C₃H₅O₂⁻(aq)
02

2. Set up an ICE table

The ICE table will allow us to track the initial concentrations, changes in concentrations, and equilibrium concentrations. | | HC₃H₅O₂ | H⁺ | C₃H₅O₂⁻ | |--------|-----------------|-----------------|------------------| | Initial| 0.100 M | 0 M | 0 M | | Change | -x M | +x M | +x M | | Equilibrium| 0.100 - x M| x M | x M |
03

3. Write the Ka expression

For the dissociation of propanoic acid, the Ka expression is: Ka = \(\frac{[H⁺][C₃H₅O₂⁻]}{[HC₃H₅O₂]}\) Since Ka is given as 1.3 x 10⁻⁵, we can substitute the equilibrium concentrations from the ICE table into the Ka expression as follows: 1.3 x 10⁻⁵ = \(\frac{x^{2}}{0.100 - x}\)
04

4. Solve for x

We will now solve for x using the quadratic formula or by making an assumption that x is very small compared to 0.100. This simplification is valid as propanoic acid is a weak acid. 1.3 x 10⁻⁵ ≈ \(\frac{x^{2}}{0.100}\) Now, solving for x: x ≈ \(0.0036\) This value of x represents the concentration of H⁺ ions at equilibrium.
05

5. Calculate the pH

To calculate the pH, we can use the formula: pH = -log[H⁺] pH = -log(0.0036) pH ≈ 2.44
06

6. Calculate the percent dissociation

The percent dissociation can be calculated as follows: Percent dissociation = \(\frac{[H⁺]_{equilibrium}}{[HC₃H₅O₂]_{initial}} \times 100\%\) Percent dissociation = \(\frac{0.0036}{0.100} \times 100\%\) Percent dissociation ≈ 3.6%
07

7. Find the concentrations of all species at equilibrium

From the ICE table, [HC₃H₅O₂] = 0.100 - x ≈ 0.100 - 0.0036 = 0.0964 M [H⁺] = x ≈ 0.0036 M [C₃H₅O₂⁻] = x ≈ 0.0036 M At equilibrium, the concentrations of the species are: [HC₃H₅O₂] ≈ 0.0964 M [H⁺] ≈ 0.0036 M [C₃H₅O₂⁻] ≈ 0.0036 M The solution to this problem is as follows: - Concentrations at equilibrium: [HC₃H₅O₂] ≈ 0.0964 M, [H⁺] ≈ 0.0036 M, and [C₃H₅O₂⁻] ≈ 0.0036 M. - pH ≈ 2.44. - Percent dissociation ≈ 3.6%.

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Most popular questions from this chapter

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

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