Question

A volume of \(18 \mathrm{ml}\) of mixture of acetic acid and sodium acetate required \(6 \mathrm{ml}\) of \(0.1 \mathrm{M}-\mathrm{NaOH}\) for neutralization of the acid and \(12 \mathrm{ml}\) of \(0.1 \mathrm{M}-\mathrm{HCl}\) reaction with salt separately. If \(\mathrm{p} K_{\mathrm{a}}\) of acetic acid is \(4.75\), what is the \(\mathrm{pH}\) of the mixture? \((\log 2=0.3)\) (a) \(5.05\) (b) \(4.45\) (c) \(4.15\) (d) \(5.35\)

Short answer

The pH of the mixture is 5.05.

Step-by-step solution

Calculate moles of acetic acid and sodium acetate

To find the moles of acetic acid and sodium acetate, use the molarity and volume of the NaOH and HCl. For acetic acid, moles are calculated by multiplying the volume of NaOH with its molarity: \(0.006 \, \text{L} \times 0.1 \, \text{M} = 0.0006 \, \text{moles} \). For sodium acetate, use the volume of HCl with molarity: \(0.012 \, \text{L} \times 0.1 \, \text{M} = 0.0012 \, \text{moles} \).

Determine the pH using the Henderson-Hasselbalch equation

The pH can be calculated using the Henderson-Hasselbalch equation: \( \text{pH} = \text{p}K_a + \log(\frac{[\text{A}^-]}{[\text{HA}]}) \). Plug in the given \( \text{p}K_a = 4.75 \), the moles of sodium acetate \(0.0012\) as \([\text{A}^-]\), and the moles of acetic acid \(0.0006\) as \([\text{HA}]\).

Calculate the ratio and apply the logarithm

Calculate the ratio of sodium acetate to acetic acid: \(\frac{0.0012}{0.0006} = 2\). Then apply the logarithm, \(\log(2) = 0.3\), so the equation becomes: \(\text{pH} = 4.75 + 0.3 = 5.05\).

Key concepts

pH Calculation

Understanding the pH of a solution is fundamental in chemistry, as it measures the acidity or basicity of that solution. It's defined by the negative logarithm of hydrogen ion concentration, expressed as the formula:
\[\begin{equation}\mathrm{pH} = -\log[H^+],\end{equation}\]
where \[\begin{equation}[H^+]}\end{equation}\]is the molarity of hydrogen ions in the solution. Molarity, which is moles of solute per liter of solution, plays a key role in calculating pH. In buffer systems such as acetic acid and sodium acetate, the pH can be accurately calculated using the Henderson-Hasselbalch equation, which provides a direct relationship between the pH and the concentration of the acid and its conjugate base in the solution. Simplifying the concept, the higher the concentration of hydrogen ions, the more acidic the solution and the lower the pH value.

Acetic acid and Sodium Acetate

Acetic acid (\[\begin{equation}CH_3COOH\end{equation}\]) is a weak acid, and when it reacts with sodium hydroxide (NaOH), the base, it forms sodium acetate (\[\begin{equation}CH_3COONa\end{equation}\]), its conjugate base. This pair, acetic acid and sodium acetate, forms a buffer solution, which resists changes in pH upon the addition of an acid or a base. When studying a buffer, we often need to know the molar amounts of the acid and its salt—information that can be deduced from neutralization reactions with standardized solutions of NaOH and HCl. By finding out how much NaOH reacts with the acetic acid, we can determine the moles of the acid. Similarly, by reacting sodium acetate with HCl, we can find out the moles of the salt.

Buffer Solution Chemistry

Buffer solutions are mixtures that contain a weak acid and its conjugate base in significant quantities. They maintain a relatively constant pH when small amounts of acids or bases are added. This principle is a cornerstone in buffer chemistry and is particularly useful in biological systems where maintaining a stable pH is critical. The Henderson-Hasselbalch equation is the mathematical expression that illustrates how buffers work. It describes the pH of a buffer solution in terms of the concentration (normally given in molarity) of the weak acid and its conjugate base. By understanding this relationship, one can predict the resulting pH after adding specific amounts of acid or base to the buffer solution.

Molarity and Neutralization

Molarity is a measure of the concentration of a solute in a solution and is defined as the number of moles of the solute divided by the volume of the solution in liters. It is used extensively in various chemical calculations, including neutralization reactions. A neutralization reaction occurs when an acid reacts with a base to produce water and a salt. Through these reactions, it's possible to calculate the molarity of acids or bases if one of them is of a known concentration. Once the moles of the reactants are known, we can understand the stoichiometry of the reaction and further calculate the amounts of reactants or products using the concept of molarity.