Question

A student intends to titrate a solution of a weak monoprotic acid with a sodium hydroxide solution but reverses the two solutions and places the weak acid solution in the buret. After \(23.75 \mathrm{~mL}\) of the weak acid solution has been added to \(50.0 \mathrm{~mL}\) of the \(0.100 \mathrm{M} \mathrm{NaOH}\) solution, the \(\mathrm{pH}\) of the resulting solution is \(10.50 .\) Calculate the original concentration of the solution of weak acid.

Short answer

The original concentration of the solution of weak acid is \(0.202 \mathrm{M}\).

Step-by-step solution

Calculate the moles of NaOH present initially

Using the given volume and concentration of NaOH, we can calculate the moles of NaOH initially present:

Step 1.1: converting volumes to Liters

Convert the given volumes from mL to L: - Volume of NaOH solution: \(50.0 \mathrm{~mL} \times \frac{1 \mathrm{~L}}{1000 \mathrm{~mL}} = 0.0500 \mathrm{~L}\) - Volume of weak acid solution: \(23.75 \mathrm{~mL} \times \frac{1 \mathrm{~L}}{1000 \mathrm{~mL}} = 0.02375 \mathrm{~L}\)

Step 1.2: Calculate the moles of NaOH

Now, we can find the initial moles of NaOH using the given concentration and the converted volume: Moles of NaOH = Molarity × Volume Moles of NaOH = \(0.100 \mathrm{M} \times 0.0500 \mathrm{~L}\) = \(0.00500 \mathrm{~mol}\)

Determine the moles of NaOH remaining

To find the moles of NaOH remaining after the addition of weak acid, we can use the given pH: pH = 10.50 implies pOH = 14 - pH = 3.50 Now, we can find the hydroxide ion concentration, \([OH^-]\), using the pOH: \([OH^-] = 10^{-pOH} = 10^{-3.50}\) The moles of remaining NaOH are equal to the moles of hydroxide ions: Moles of remaining NaOH = \([OH^-] \times V_{total}\) Moles of remaining NaOH = \(10^{-3.50} \times (0.0500 \mathrm{~L} + 0.02375 \mathrm{~L}) = 1.971 \times 10^{-4} \mathrm{~mol}\)

Calculate the moles of weak acid added

The moles of weak acid added, which reacted with NaOH can be found by subtracting moles of remaining NaOH from the initial moles of NaOH: Moles of weak acid added = Moles of initial NaOH - Moles of remaining NaOH Moles of weak acid added = \(0.00500 \mathrm{~mol} - 1.971 \times 10^{-4} \mathrm{~mol} = 4.803 \times 10^{-3} \mathrm{~mol}\)

Find the concentration of the weak acid

Finally, we can find the concentration of the weak acid using the moles of weak acid added and the volume of the weak acid solution: Concentration of weak acid = Moles of weak acid added / Volume of weak acid solution Concentration of weak acid = \(4.803 \times 10^{-3} \mathrm{~mol} / 0.02375 \mathrm{~L} = 0.202 \mathrm{M}\) So the original concentration of the weak acid solution is \(0.202 \mathrm{M}\).

Key concepts

Understanding Weak Acids

Weak acids are acids that do not completely dissociate in water. This means that only a portion of the acid molecules release hydrogen ions ( H^+ ), leaving a mix of ionized and non-ionized molecules in solution. As a result, the concentration of H^+ ions in weak acids is typically lower compared to strong acids, which dissociate almost completely.

Weak acids have unique properties:

• **Partial Dissociation**: Unlike strong acids, weak acids do not give up all hydrogen ions in solution.
• **Chemical Equilibrium**: Weak acids engage in a dynamic equilibrium between the ionized and unionized forms.
• **Low pKa Values**: They have higher pKa values which reflect their lower tendency to release H^+ ions.

These properties make weak acids less reactive but useful in many titration experiments where gradual changes in pH are desirable.

Role of Sodium Hydroxide in Titration

Sodium hydroxide (NaOH) is a strong base commonly used in titrations. When conducting acid-base titrations, NaOH acts as the titrant, meaning it is the solution of known concentration used to determine the concentration of an unknown acid or base.

Sodium hydroxide is effective in titrations because:

• **Strong Base**: NaOH completely dissociates in water, providing a reliable and consistent source of hydroxide ions ( OH^- ).
• **Mixes Well**: It readily reacts with hydrogen ions from acids to form water, an essential part of reaching the equivalence point.
• **Easy to Prepare**: Its solid form dissolves easily in water, creating solutions of precise molarity.

In titration exercises, the reaction of NaOH with a weak acid helps us determine the acid's concentration by carefully measuring how much NaOH is needed to reach a specific endpoint indicated by a noticeable change in pH.

Calculating pH in Titration

Calculating pH during a titration involves understanding both the contribution of the acid and base in the reaction. The pH is a measure of the acidity or basicity of a solution, and it represents the concentration of hydrogen ions.

To find the pH:

• **Identify Contributions**: Both the remaining NaOH and the partially neutralized acid contribute to the pH.
• **Use the pH Formula**: pH is calculated using \( pH = -\log[H^+] \) for the resultant acidic or basic environment.
• **Find the pOH**: Sometimes, determining \( pOH \) can be easier, especially with bases. This involves \( pOH = -\log[OH^-] \), and then \( pH = 14 - pOH \) to find the pH.

In cases where a weak acid and a strong base are involved, the predominant species will affect the pH. Therefore, calculation often requires understanding the equilibrium state achieved in the solution.

Understanding Acid-Base Reactions

Acid-base reactions are fundamental chemical processes where acids donate protons ( H^+ ), and bases accept them. In this context, they form products like water and salt in total or partial exchanges.

Key aspects of acid-base reactions include:

• **Proton Exchange**: Acids donate H^+ ions to bases in a typical reaction.
• **Equivalence Point**: In a titration, this is where the number of moles of acid equals the moles of base, resulting in neutralization.
• **pH Change**: Monitoring pH helps identify the endpoint where the reaction has reached completion.

These reactions are integral to many laboratory practices and are used to determine concentrations, understand solubility, and explore chemical properties of substances.