Question

A solution is made by adding \(50.0 \mathrm{mL}\) of \(0.200 \mathrm{M}\) acetic acid \(\left(K_{\mathrm{a}}=1.8 \times 10^{-5}\right)\) to \(50.0 \mathrm{mL}\) of \(1.00 \times 10^{-3} \mathrm{M} \mathrm{HCl}\) a. Calculate the \(p\) H of the solution. b. Calculate the acetate ion concentration.

Short answer

a. To calculate the pH of the solution, first, find the equilibrium concentrations of the species. Then, apply the weak acid equilibrium expression to find the concentration of H\(^{+}\) ions in the solution: H\(^{+}\) ion concentration = \(\frac{[\mathrm{CH_3COOH}]K_{a}}{[\mathrm{CH_3COO^-}]} =\frac{[(1.0\times10^{-2} - 5.0 \times 10^{-5})\;\mathrm{mol}/\mathrm{L}(1.8 \times 10^{-5})]}{ (5.0 \times 10^{-5}\;\mathrm{mol}/\mathrm{L})}\) Finally, calculate the pH using the formula pH = -log10(H\(^{+}\) ion concentration). b. The acetate ion concentration is directly obtained from the moles of acetate ions formed divided by the total volume of the solution: Acetate ion concentration = \(\frac{5.0 \times 10^{-5} \;\mathrm{mol}}{1.0 \times 10^{-1} \;\mathrm{L}}\).

Step-by-step solution

Determine the moles of each acid in the solution.

We are given the volumes and concentrations of the acetic acid and HCl solutions. We can use these values to calculate the moles of each acid in the solution. Moles of acetic acid = volume × concentration = \(50.0 \times 10^{-3} \mathrm{L} \times 0.200 \mathrm{M}\) = \(1.0 \times 10^{-2} \mathrm{mol} \) Moles of HCl = volume × concentration = \(50.0 \times 10^{-3} \mathrm{L} \times 1.00 \times 10^{-3} \mathrm{M}\) = \(5.0 \times 10^{-5} \mathrm{mol} \)

Find the equilibrium concentrations of the species

The HCl will react completely with the acetic acid to form acetate ions and water, since it is a strong acid. \( \begin{aligned} \mathrm{HCl} + \mathrm{CH_3COOH} &\rightarrow \mathrm{CH_3COO^-} + \mathrm{H_2O} \end{aligned} \) Moles of remaining acetic acid at equilibrium = moles of initial acetic acid - moles of HCl = \((1.0 \times 10^{-2}-5.0 \times 10^{-5})\;\mathrm{mol}\) Moles of formed acetate ions = moles of consumed HCl = \(5.0 \times 10^{-5} \mathrm{mol} \) Total volume of the solution = initial volume of acetic acid + initial volume of HCl = \(50.0+50.0\) mL = \(100\) mL = \(1.0 \times 10^{-1} \mathrm{L} \) Concentration of acetic acid at equilibrium = moles of remaining acetic acid / total volume of the solution = \[\frac{(1.0\times10^{-2}-5.0 \times 10^{-5})\;\mathrm{mol}}{1.0\times10^{-1}\mathrm{L}}\] Concentration of acetate ions at equilibrium = moles of acetate ions / total volume of the solution = \( \frac{5.0 \times 10^{-5}\; \mathrm{mol}}{1.0 \times 10^{-1}\mathrm{L}}\)

Calculate the pH of the solution

We apply the weak acid equilibrium expression to find the concentration of H\(^{+}\) ions in the solution: \(K_{a}\) = \(\frac{[\mathrm{CH_3COO^-}][\mathrm{H^+}]}{[\mathrm{CH_3COOH}]}\) Let the equilibrium concentration of H\(^{+}\) ions be represented by x. Since the reaction does not consume any significant amount of acetate ions, we can assume that their concentration stays constant. \(1.8 \times 10^{-5} = \frac{[\mathrm{CH_3COO^-}][x]}{[\mathrm{CH_3COOH}]}\) Solve for x: \[x = \frac{[\mathrm{CH_3COOH}]K_{a}}{[\mathrm{CH_3COO^-}]}\] Now we can substitute in the concentrations of the species at equilibrium and calculate the H\(^{+}\) ion concentration: H\(^{+}\) ion concentration = \[x = \frac{[(1.0\times10^{-2} - 5.0 \times 10^{-5})\;\mathrm{mol}/\mathrm{L}(1.8 \times 10^{-5})]}{ (5.0 \times 10^{-5}\;\mathrm{mol}/\mathrm{L})}\] Next, calculate the pH of the solution: pH = -log10(H\(^{+}\) ion concentration)

Calculate the acetate ion concentration

The concentration of acetate ions can be directly obtained from Step 2 as the moles of acetate ions formed divided by the total volume of the solution. Acetate ion concentration = \(\frac{5.0 \times 10^{-5} \;\mathrm{mol}}{1.0 \times 10^{-1} \;\mathrm{L}}\) This gives us the final answer for the acetate ion concentration.

Key concepts

Acetic Acid

Acetic acid, commonly known as ethanoic acid, is a weak acid found in vinegar, responsible for its sour taste and pungent smell. In its chemical formula, it is represented as \( \text{CH}_3\text{COOH} \). Because of its carboxylic acid group, acetic acid can release a proton (\( \text{H}^+ \)) leading to the formation of its conjugate base, acetate ion (\( \text{CH}_3\text{COO}^- \)).
- Acetic acid partially dissociates in aqueous solutions, making it a weak acid with a specific acid dissociation constant, \( K_a \). In this example, \( K_a \) is given as \( 1.8 \times 10^{-5} \), indicating its strength in donating protons.- Being a weak acid, acetic acid forms equilibrium in water between its undissociated form and its dissociated ions. This equilibrium state is central to understanding buffer solutions and calculating pH.
Knowing about acetic acid’s behavior in water is key for predicting how it will act when mixed with other substances, such as hydrochloric acid in this problem.

Buffer Solution

A buffer solution consists of a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid, which helps resist changes in pH upon the addition of small amounts of acids or bases.
- In the given example, when acetic acid is mixed with hydrochloric acid (HCl), the HCl, being a strong acid, completely dissociates in water, freeing up \( \text{H}^+ \) ions which react with the acetic acid.- This reaction results in the formation of acetate ions from the weak acetic acid, transforming the solution into a buffer system. Even with the addition of HCl, the buffer solution maintains a relatively constant pH, which is useful in various chemical and biological processes.
The ability of buffer solutions to maintain pH even when acids or bases are added is an essential property exploited in many practical applications, from laboratories to living organisms.

pH Calculation

To determine the acidity or basicity of a solution, we calculate its pH, which is a logarithmic scale representation of the hydrogen ion concentration (\( [\text{H}^+] \)). The equation for pH is given by \( \text{pH} = -\log_{10}([\text{H}^+]) \).
- In buffered systems, such as the one with acetic acid and acetate ion, pH calculations help evaluate the buffer's capacity to maintain a stable pH.- Using the equilibrium concentrations of acetic acid and its acetate ion, the concentration of hydrogen ions \( [\text{H}^+] \) can be calculated utilizing the given \( K_a \) expression. Once \( [\text{H}^+] \) is known, simply taking the negative logarithm provides the pH of the solution.
Understanding pH is critical because it affects chemical reactions, enzyme activities, and overall stability of biological systems, thus making accurate calculations indispensable.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a practical formula used to estimate the pH of a buffer solution. It is formulated as follows:\[\text{pH} = \text{pK}_a + \log_{10}\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)\]- Here, \( \text{pK}_a \) is the negative logarithm of the acid dissociation constant (\( K_a \)) and \([\text{A}^-]\) and \([\text{HA}]\) represent the concentrations of the conjugate base and the weak acid.- This equation simplifies the calculation of pH in buffer solutions consisting of weak acids and their salts.- In the context of the given problem, after calculating the equilibrium concentrations, the Henderson-Hasselbalch equation provides an easy way to find the pH by plugging in the values of \( \text{CH}_3\text{COO}^- \) and \( \text{CH}_3\text{COOH} \). The Henderson-Hasselbalch equation is particularly useful for preparing solutions at a desired pH and understanding buffer capacity in laboratory and industrial applications.