Question

\(0.001 \mathrm{~mol}\) of \(\mathrm{NaOH}\) is added to \(100 \mathrm{ml}\). of a solution that is \(0.5 \mathrm{M} \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) and \(0.5 \mathrm{M} \mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\). Determine the \(\mathrm{pH}\) of this Solution; the \(\mathrm{K}_{\text {diss }}\) of acetic acid is \(1.8 \times 10^{-5}\).

Short answer

The pH of the solution after adding 0.001 mol of NaOH to the given solution is approximately 4.75.

Step-by-step solution

Identify the main species in the solution

Before we begin, we need to recognize the species in the solution. We have acetic acid (HC2H3O2) and sodium acetate (NaC2H3O2) as the main species, and we are adding sodium hydroxide (NaOH).

Calculate moles of HC2H3O2 and NaC2H3O2 in the solution

We are given the volume of the solution as 100 ml and the concentrations of HC2H3O2 and NaC2H3O2 as 0.5 M. To find the moles of these species in the solution, we can use the formula: Moles = Molarity × Volume (in liters) For HC2H3O2: Moles = (0.5 mol/L) × (0.100 L) = 0.050 mol For NaC2H3O2: Moles = (0.5 mol/L) × (0.100 L) = 0.050 mol

Calculate the change in moles due to the reaction with NaOH

We are given that 0.001 mol of NaOH is added to the solution. When NaOH reacts with HC2H3O2, it will consume one mole of HC2H3O2 and produce one mole of NaC2H3O2 for every mole of NaOH: NaOH + HC2H3O2 -> H2O + NaC2H3O2 Since we have 0.001 mol of NaOH and 0.050 mol of HC2H3O2, the reaction will proceed until all the NaOH is consumed. The change in moles of HC2H3O2 and NaC2H3O2 will be as follows: HC2H3O2: 0.050 mol - 0.001 mol = 0.049 mol NaC2H3O2: 0.050 mol + 0.001 mol = 0.051 mol

Calculate the new concentrations of HC2H3O2 and NaC2H3O2 after reaction

We need to calculate the new concentrations of HC2H3O2 and NaC2H3O2 after the reaction with NaOH. The volume of the solution remains 100 ml (0.100 L). So, we can calculate the new concentrations using the formula: Concentration = Moles ÷ Volume (in liters) For HC2H3O2: Concentration_after_reaction = (0.049 mol) ÷ (0.100 L) = 0.49 M For NaC2H3O2: Concentration_after_reaction = (0.051 mol) ÷ (0.100 L) = 0.51 M

Use the Henderson-Hasselbalch equation to determine the pH

Now, we will use the Henderson-Hasselbalch equation to determine the pH of the solution after adding NaOH: pH = pKa + log10([Conjugate Base] ÷ [Acid]) First, we need to determine the pKa from the given Kdiss of acetic acid: Kdiss = 1.8 x 10^(-5) pKa = -log10(Kdiss) = -log10(1.8 x 10^(-5)) ≈ 4.74 Now we can plug in the values into the Henderson-Hasselbalch equation: pH = 4.74 + log10(0.51 ÷ 0.49) ≈ 4.74 + 0.014 ≈ 4.75

Final Answer

The pH of the solution after adding 0.001 mol of NaOH to the given solution is approximately 4.75.

Key concepts

Henderson-Hasselbalch Equation

Understanding the Henderson-Hasselbalch equation is essential for students tackling buffer chemistry problems. Primarily, this equation provides a direct relationship between the pH of a buffer solution and the ratio of the concentration of the conjugate base to its corresponding acid. It’s derived from the equilibrium constant expression for a weak acid and is formulated as:

\[\text{pH} = \text{pKa} + \log\left(\frac{[\text{Conjugate Base}]}{[\text{Acid}]}\right)\]

This equation assumes that the concentration of the acid and its conjugate base, and the acidic dissociation constant (\(\text{Ka}\)) are known. The \(\text{pKa}\) is the negative logarithm of \(\text{Ka}\) (\(\text{pKa} = -\log(\text{Ka})\)). For instance, with acetic acid as the weak acid and acetate as the conjugate base, this equation allows prediction of the resulting \(\text{pH}\) change when a strong base like NaOH is added to the buffer solution. The key is understanding that adding NaOH shifts the ratio by converting some of the acid into its conjugate base.

Buffer Solution Chemistry

Buffer solutions are fascinating due to their ability to resist changes in pH when small amounts of acids or bases are added. This quality is harnessed in various applications, including biochemical labs and industry where pH control is crucial.

Buffers are composed of a weak acid and its conjugate base or a weak base and its conjugate acid. The magic of a buffer’s pH stability lies in the weak acid-base equilibrium that quickly re-establishes when disturbed by the addition of stronger acids or bases. The presence of both components ensures that any added hydroxide ions (\(\text{OH}^-\)) react with the acid, while any added hydrogen ions (\(\text{H}^+\)) react with the conjugate base, maintaining the solution's pH with minimal change.

In the provided exercise, the buffer is initially composed of equal concentrations of acetic acid and sodium acetate, manifesting its optimal buffering capacity around the acetic acid pKa value.

Acetic Acid and Acetate Buffer

A buffer composed of acetic acid (\(\text{CH}_3\text{COOH}\) or \(\text{HC}_2\text{H}_3\text{O}_2\)) and its conjugate base, acetate (\(\text{CH}_3\text{COO}^-\) or \(\text{C}_2\text{H}_3\text{O}_2^-\)), exemplifies a classic weak acid-strong base buffer system.

When a small amount of a strong base like NaOH is added to this buffer, NaOH dissociates completely, raising the concentration of acetate as it reacts with acetic acid to form water and the acetate ion. This reaction slightly shifts the equilibrium but doesn't cause a significant pH change because of the buffering action. Here the relative amounts of acetic acid to acetate after the reaction determine the new pH using the Henderson-Hasselbalch equation.

The exercise solution elegantly demonstrates this by adding a known quantity of NaOH to the buffer and calculating the new conjugate pair concentrations to determine the final pH. The teaching point here is the demonstration of the buffer's capacity to neutralize added base and the calculation of subsequent pH changes.