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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: [HC鈧僅鈧匫鈧俔 鈮 0.0964 M, [H鈦篯 鈮 0.0036 M, and [C鈧僅鈧匫鈧傗伝] 鈮 0.0036 M. The pH of this solution is approximately 2.44, and the percent dissociation of propanoic acid is 鈮 3.6%.

Step by step solution

01

1. Write the dissociation equation

The dissociation equation for propanoic acid is: HC鈧僅鈧匫鈧(aq) 鈬 H鈦(aq) + C鈧僅鈧匫鈧傗伝(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. | | HC鈧僅鈧匫鈧 | H鈦 | C鈧僅鈧匫鈧傗伝 | |--------|-----------------|-----------------|------------------| | 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 = \(\frac{[H鈦篯[C鈧僅鈧匫鈧傗伝]}{[HC鈧僅鈧匫鈧俔}\) Since Ka is given as 1.3 x 10鈦烩伒, we can substitute the equilibrium concentrations from the ICE table into the Ka expression as follows: 1.3 x 10鈦烩伒 = \(\frac{x^{2}}{0.100 - x}\)
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鈦烩伒 鈮 \(\frac{x^{2}}{0.100}\) Now, solving for x: x 鈮 \(0.0036\) This value of x represents the concentration of H鈦 ions at equilibrium.
05

5. Calculate the pH

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

6. Calculate the percent dissociation

The percent dissociation can be calculated as follows: Percent dissociation = \(\frac{[H鈦篯_{equilibrium}}{[HC鈧僅鈧匫鈧俔_{initial}} \times 100\%\) Percent dissociation = \(\frac{0.0036}{0.100} \times 100\%\) Percent dissociation 鈮 3.6%
07

7. Find the concentrations of all species at equilibrium

From the ICE table, [HC鈧僅鈧匫鈧俔 = 0.100 - x 鈮 0.100 - 0.0036 = 0.0964 M [H鈦篯 = x 鈮 0.0036 M [C鈧僅鈧匫鈧傗伝] = x 鈮 0.0036 M At equilibrium, the concentrations of the species are: [HC鈧僅鈧匫鈧俔 鈮 0.0964 M [H鈦篯 鈮 0.0036 M [C鈧僅鈧匫鈧傗伝] 鈮 0.0036 M The solution to this problem is as follows: - Concentrations at equilibrium: [HC鈧僅鈧匫鈧俔 鈮 0.0964 M, [H鈦篯 鈮 0.0036 M, and [C鈧僅鈧匫鈧傗伝] 鈮 0.0036 M. - pH 鈮 2.44. - Percent dissociation 鈮 3.6%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

pH Calculation
The pH of a solution is a measure of its acidity or basicity and is calculated from the concentration of hydrogen ions \([H^+]\). When dealing with weak acids like propanoic acid, it鈥檚 essential to first find the equilibrium concentration of \([H^+]\).
Once you have it, you use the formula \( ext{pH} = -\log[H^+]\). For our specific example:
\[\text{pH} = -\log(0.0036)\]
This results in a pH of approximately 2.44, indicating an acidic solution. The negative logarithm in the formula gives you the power of 10, making it a simple yet efficient way to express the hydrogen ion concentration.
ICE Table
An ICE table is a systematic way to organize data for reactions in equilibrium. Here, 'I' stands for initial concentration, 'C' for change in concentration, and 'E' for equilibrium concentration.
This tool helps in visualizing how concentrations shift as a chemical reaction proceeds toward equilibrium.
  • Initial Concentration (I): Start with 0.100 M for propanoic acid and 0 M for products (H鈦 and C鈧僅鈧匫鈧傗伝).
  • Change (C): As the reaction proceeds, the amount of acid decreases by \-x\ M, while both H鈦 and C鈧僅鈧匫鈧傗伝 increase by \+x\ M.
  • Equilibrium (E): Use the expressions \(0.100 - x\) for the acid, and x for the ions.
This concise table simplifies complex equilibrium problems helping you to define the unknowns clearly.
Ka Expression
The acid dissociation constant, \(K_a\), tells us how well an acid can donate a proton. It鈥檚 essential in calculating equilibrium concentrations. For propanoic acid, the expression is:
\[K_a = \frac{[H^+][C鈧僅鈧匫鈧俕-]}{[HC鈧僅鈧匫鈧俔}\]
With \(K_a = 1.3 \times 10^{-5}\). By substituting the equilibrium concentrations (from the ICE table), we can write:
\[1.3 \times 10^{-5} = \frac{x^2}{0.100 - x}\]
Simplifying often assumes \(x \approx 0.0036\), as x is small compared to the initial acid concentration (a valid approximation for weak acids), converting the equation to:
\[1.3 \times 10^{-5} \approx \frac{x^2}{0.100}\]
This simplifies to solve for \(x\), the concentration of H鈦 at equilibrium.
Percent Dissociation
Percent dissociation shows how much of the initial acid dissociates into ions, an important aspect to consider in equilibrium studies. Use the formula:
\[\text{Percent Dissociation} = \frac{[H^+]_{\text{equilibrium}}}{[HC鈧僅鈧匫鈧俔_{\text{initial}}} \times 100\%\]
For propanoic acid, this becomes:
\[\frac{0.0036}{0.100} \times 100\% \approx 3.6\%\]
This low percentage confirms that propanoic acid is indeed a weak acid, as only a small fraction dissociates in solution. This calculation helps in comparing the strength of different acids and is a crucial part of equilibrium analysis.

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Most popular questions from this chapter

For the following, mix equal volumes of one solution from Group I with one solution from Group II to achieve the indicated \(\mathrm{pH}\). Calculate the \(\mathrm{pH}\) of each solution. Group I: \(0.20 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}, 0.20 \mathrm{MHCl}, 0.20 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{Cl}, 0.20\) \(M\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3} \mathrm{NHCl}\) Group II: \(0.20 \mathrm{M} \mathrm{KOI}, 0.20 \mathrm{M} \mathrm{NaCN}, 0.20 \mathrm{M} \mathrm{KOCl}, 0.20 \mathrm{M}\) \(\mathrm{NaNO}_{2}\) a. the solution with the lowest \(\mathrm{pH}\) b. the solution with the highest \(\mathrm{pH}\) c. the solution with the \(\mathrm{pH}\) closest to \(7.00\)

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?

Write out the stepwise \(K_{\mathrm{a}}\) reactions for the diprotic acid \(\mathrm{H}_{2} \mathrm{SO}_{3}\).

Which of the following represent conjugate acid-base pairs? For those pairs that are not conjugates, write the correct conjugate acid or base for each species in the pair. a. \(\mathrm{H}_{2} \mathrm{O}, \mathrm{OH}\) c. \(\mathrm{H}_{3} \mathrm{PO}_{4}, \mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) b. \(\mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{SO}_{4}^{2-}\) d. \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}, \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}\)

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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