Question

An acetic acid-sodium acetate buffer can be prepared by the reaction \(\mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{H}_{3} \mathrm{O}^{+} \longrightarrow \mathrm{CH}_{3} \mathrm{COOH}+\mathrm{H}_{2} \mathrm{O}\) (From \(\mathrm{NaCH}_{3} \mathrm{COO}\) )(From HCl) (a) If \(12.0 \mathrm{g} \mathrm{NaCH}_{3} \mathrm{COO}\) is added to \(0.300 \mathrm{L}\) of 0.200 M HCl, what is the pH of the resulting solution? (b) If \(1.00 \mathrm{g} \mathrm{Ba}(\mathrm{OH})_{2}\) is added to the solution in part (a), what is the new pH? (c) What is the maximum mass of \(\mathrm{Ba}(\mathrm{OH})_{2}\) that can be neutralized by the buffer solution of part (a)? (d) What is the pH of the solution in part (a) following the addition of \(5.50 \mathrm{g} \mathrm{Ba}(\mathrm{OH})_{2} ?\)

Short answer

(a) The pH of the resulting solution is 5.01. (b) The new pH after adding 1.00 g of Ba(OH)2 is 5.05. (c) The maximum mass of Ba(OH)2 that can be neutralized by the buffer solution in part (a) is 5.14 g. (d) The pH of the solution after adding 5.50 g of Ba(OH)2 is 12.12.

Step-by-step solution

Determine the moles of NaCH3COO and HCl.

The number of moles of NaCH3COO can be calculated with the formula n = m / M, where n is the number of moles, m is the mass of the substance, and M is the molecular mass of the substance. The mass of NaCH3COO is given as 12.0 g, and the molecular mass is approximately 82.0 g/mol. Thus, the moles of NaCH3COO are \( \frac{12.0 g }{82.0 g/mol} \approx 0.146 mol\). Then, calculate the moles of HCl. HCl is in a 0.200 M solution of 0.300 L, so its moles are: \( (0.200 mol/L) * 0.300 L = 0.060 mol \).

Determine the reaction at equilibrium.

The moles of NaCH3COO and HCl will react to form CH3COOH and H2O. This reaction will occur simultaneously until one of the two initial reactants is completely consumed. Since there are fewer moles of HCl than NaCH3COO, HCl will be completely consumed. As HCl (0.060 mol) reacts, it reduces the amount of NaCH3COO by 0.060 mol and increases the amount of CH3COOH by 0.060 mol. Calculate the equilibrium amounts: NaCH3COO is 0.146 mol - 0.060 mol = 0.086 mol; CH3COOH is 0 mol + 0.060 mol = 0.060 mol.

Calculate the pH of the solution in part (a)

Use the Henderson-Hasselbalch equation to find the pH: \( pH = pKa + log\frac{{[A^-]}}{{[HA]}} \). For acetic acid, pKa is 4.76. Substitute the values into the equation: \( pH = 4.76 + log\frac{{0.086}}{{0.060}} \approx 5.01 \). Thus, the pH of the solution is around 5.01.

Calculate the moles of Ba(OH)2 and the new pH in part (b)

First, calculate the total moles of Ba(OH)2. The molecular mass of Ba(OH)2 is approximately 171.3 g/mol. Thus, the moles of Ba(OH)2 are \( \frac{1.00 g}{171.3 g/mol} \approx 0.0058 mol\). Since Ba(OH)2 can produce 2 OH- ions for every molecule, there will be twice the number of OH- ions in the solution. Now, OH- ions would react with CH3COOH from the buffer, producing CH3COO- and H2O. At equilibrium, Ba(OH)2 is entirely dissociated and we are left with 0.060 mol - 0.0058 mol ≈ 0.054 mol of CH3COOH, 0.086 mol - 0.0058 mol ≈ 0.080 mol of CH3COO-. Use the Henderson-Hasselbalch equation again to find the new pH, which should be approximately 5.05.

Calculate the maximum mass of Ba(OH)2 that can be neutralized in part (c)

The buffer can neutralize the Ba(OH)2 until the CH3COOH is entirely depleted. By stoichiometry, each mole of CH3COOH in solution can neutralize an equivalent mole of OH-. Since each mole of Ba(OH)2 produces 2 moles of OH-, the total amount of Ba(OH)2 that can be neutralized by the solution is 1/2 of the amount of CH3COOH initially in solution: \( \frac{0.060 mol}{2}=0.030 mol \). So, the maximum mass of Ba(OH)2 is \( 0.030 mol * 171.3 g/mol = 5.14 g\).

Determine the pH of the solution when 5.5 g Ba(OH)2 is added in part (d)

First, calculate the moles of Ba(OH)2: \( \frac{5.5 g}{171.3 g/mol } = 0.032 mol\). It can be seen that the moles of Ba(OH)2 added exceeds the buffer capacity calculated in the previous step, meaning it will not all be neutralized by the buffer. Ba(OH)2 reacts with the buffer to produce the maximum capability of the buffer. The remaining Ba(OH)2 will then dissociate into Ba2+ and OH- ions, and the excess OH- ions will result in a basic solution. Calculate the moles of OH- ions in excess after the buffer capacity is exceeded: \( 0.032 mol *2 - 0.060 mol = 0.004 mol\). Then, calculate the concentration of OH- ions: \( 0.004 mol / 0.300 L = 0.0133 M\). From this, calculate the pOH of the solution: \( -log(0.0133) = 1.88 \). Convert pOH to pH by subtracting from 14 to get the final pH = 12.12

Key concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a fundamental formula used to estimate the pH of buffer solutions, which are systems that maintain a relatively constant pH when small amounts of acid or base are added. Buffer solutions typically consist of a weak acid and its conjugate base. The general form of the Henderson-Hasselbalch equation is:

\[\begin{equation}pH = pKa + log\frac{{[A^-]}}{{[HA]}}\end{equation}\]
where:\begin{itemize}\item \(pH\) is the acidity of the solution,\item \(pKa\) represents the acid dissociation constant, indicating the strength of the weak acid,\item \([A^-]\) is the concentration of the conjugate base, and\item \([HA]\) is the concentration of the weak acid.\end{itemize}
The logarithmic term in this equation expresses the ratio of the dissociated ions of the weak acid (base form) to the undissociated form (acidic form). The beauty of the Henderson-Hasselbalch equation lies in its simplicity; it allows the pH to be calculated easily, assuming the concentrations of the acid and conjugate base are known and the solution behaves ideally. It is particularly useful in the realm of biochemistry, where many reactions occur in buffered environments such as human blood. To solve for pH when given the masses of a weak acid and its salt, as in our original exercise, one must convert those masses to moles and plug them into the equation, being careful to account for volume differences if necessary.

pH Calculation

pH calculation is an essential process to determine the acidity or basicity of a solution. pH is defined as the negative logarithm (base 10) of the concentration of hydrogen ions (\(H^+\)) in a solution. Mathematically, the pH is calculated using the formula:

\[\begin{equation}pH = -log[H^+]\end{equation}\]
This formula reveals the inverse logarithmic relationship between \(H^+\) ion concentration and pH: as \(H^+\) concentration increases, indicating a more acidic solution, the pH value decreases. Conversely, a decrease in \(H^+\) concentration, indicating a more basic solution, results in an increase in pH. The pH scale generally ranges from 0 to 14, with 7 as the neutral point. Solutions with a pH less than 7 are acidic, while those with a pH greater than 7 are basic or alkaline. An important aspect touched upon in the exercise is understanding that the addition of a strong base, such as \(Ba(OH)_2\), can shift the balance between the weak acid and its conjugate base in a buffer solution. In the given example, the moles of base added can convert the remaining weak acid into its conjugate base, potentially altering the pH, which can be calculated using the adjusted concentrations with the Henderson-Hasselbalch equation.

Chemical Equilibrium

Chemical equilibrium is a state in which the rate of the forward reaction is equal to the rate of the reverse reaction, resulting in no overall change in the concentrations of reactants and products over time. In the context of buffer systems, equilibrium is vital for maintaining a consistent pH level. When a weak acid (HA) is present in a solution with its conjugate base (A-), an equilibrium is established as represented by the chemical equation:

\[\begin{equation}HA \longleftrightarrow A^- + H^+\end{equation}\]
This equilibrium is described quantitatively by the acid dissociation constant (Ka), which is a measure of the strength of the acid and its tendency to donate a proton. At equilibrium, the concentration of the products multiplied together, divided by the concentration of the reactants, equals the constant:\begin{itemize}\item \[\begin{equation}Ka = \frac{{[A^-][H^+]}}{{[HA]}}\end{equation}\]\end{itemize}
In our exercise, when the reaction includes the addition of a strong acid or base, it can disturb the established equilibrium. However, the buffer solution resists changes in pH by shifting its equilibrium position through Le Châtelier's principle: the system adjusts to partially counteract the change. This shifting of balances is critical in determining how much additional acid or base the buffer can neutralize, as explored in parts (c) and (d) of the original exercise. Always consider the mole ratio of the reactants and products when dealing with a buffer's capacity to neutralize added acids or bases to maintain chemical equilibrium.