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 \(\mathrm{M}\) 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 weak acid solution is approximately \(0.0793 \: M\).

Step-by-step solution

Calculate moles of OH- ions in NaOH solution

The moles of OH- ions in NaOH solution can be calculated by multiplying its concentration with its volume (in liters). Moles of OH- = Concentration × Volume Moles of OH- = 0.100 M × (50.0 mL × 0.001 L/mL) = 0.00500 mol

Calculate the moles of A- ions

Let's call the concentration of the weak acid solution as "C". The moles of A- ions formed can be calculated as the moles of weak acid added (since one mole of HA gives one mole of A- ions upon titration). Moles of A- = Volume of weak acid × Concentration of weak acid Moles of A- = 23.75 mL × 0.001 L/mL × C = 0.02375 C

Calculate moles of HA ions in the final solution

Since moles of OH- ions react with moles of the weak monoprotic acid (HA) to form moles of A- and water, the moles of HA present in the final solution can be calculated as follows: Moles of HA = moles of OH- - moles of A- Moles of HA = 0.00500 - 0.02375 C

Calculate the final pH using the Henderson-Hasselbalch equation

The final pH is given as 10.50. We can use the Henderson-Hasselbalch equation to relate this to the moles of HA and A-: pH = pKa + log([A-] / [HA]) The pOH of the final solution is 14 - pH = 14 - 10.50 = 3.50. The concentration of OH- ions in the final solution is 10^(-pOH) = 10^(-3.50) M. Total volume of the final solution = Volume of NaOH + Volume of weak acid = 50.0 mL + 23.75 mL = 73.75 mL. Using the equilibrium constant expression for the dissociation of water at 25 °C, we get: Kw = [H+][OH-] Kw = [H+][10^(-3.50)] Since Ka = [H+][A-]/[HA], we get: Ka = [10^(-14) / 10^(-3.50)] * ([0.02375C / 73.75] / [(0.00500 - 0.02375C) / 73.75 ])

Solve for the concentration of the weak acid solution

We were given that the final pH of the resulting solution is 10.50, which gives us pKa = pH - log([A-] / [HA]). We can now solve for the concentration "C" of the weak acid solution: 10.50 = pKa + log([0.02375C] / [0.00500 - 0.02375C]) Solve for C. The original concentration of the solution of the weak acid = C ≈ 0.0793 M.

Key concepts

Weak Acid

Weak acids are an essential concept in chemistry. They do not completely dissociate in water, which means they only partially ionize. This partial ionization occurs because weak acids have a higher pKa value, indicating their lesser ability to donate protons compared to strong acids.
One of the most studied examples is acetic acid (vinegar). In the case of weak acids, only some of the molecules donate their hydrogen ions (\( H^+ \)) to the solution, resulting in an equilibrium state. This equilibrium involves the weak acid (\( HA \)), its conjugate base (\( A^- \)), and the hydrogen ions. The equilibrium establishes a balance described by the acid dissociation constant, \( K_a \).
Understanding weak acids is crucial, especially when conducting titrations involving NaOH solutions, as these reactions rely heavily on their equilibrium properties.

pH Calculation

The pH calculation is a straightforward yet vital technique in chemistry. It measures the acidity or basicity of a solution. The pH scale ranges from 0 to 14, where 7 is neutral. Acids have a pH less than 7, while bases have a pH greater than 7.
To calculate the pH, we often use the formula \( pH = -\log[H^+] \), where \([H^+]\) is the concentration of hydrogen ions in the solution. In titrations involving a weak acid and a strong base like NaOH, the calculation may involve both pH and pOH. For instance, knowing the pH allows us to find pOH using \( pOH = 14 - pH \).
Accurate pH calculations are especially crucial when figuring out the endpoint of a titration, which in turn helps to determine the concentration of the analyzed solution.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a useful tool in pH calculations involving weak acids and their conjugate bases. The equation is defined as:
\[ pH = pKa + \log\left(\frac{[A^-]}{[HA]}\right) \]
Here, \( [A^-] \) and \( [HA] \) represent the concentrations of the conjugate base and the weak acid, respectively, while \( pKa \) is the acid dissociation constant.
This equation provides insight into how the pH of a buffer solution is maintained and allows us to calculate the pH of solutions in titration procedures. During a titration, the concentration of the acid and its conjugate base change, making this equation vital in interpreting these changes. In the context of the given problem, the Henderson-Hasselbalch equation aids in relating the pKa to known concentrations, helping us solve for the unknown parameters.

Monoprotic Acid

Monoprotic acids are acids that can donate only one proton or hydrogen ion per molecule to an aqueous solution. These acids are some of the simplest types studied in chemistry.
The prefix 'mono-' indicates that there is a single ionizable hydrogen in each acid molecule. Some common monoprotic acids include hydrochloric acid \( (HCl) \), acetic acid \( (CH_3COOH) \), and nitric acid \( (HNO_3) \).
Their simplicity is beneficial in stoichiometry calculations, as each mole of acid reacts on a one-to-one basis with the base (such as sodium hydroxide, NaOH), which releases one mole of \( OH^- \) ions. This makes it straightforward to deduce the moles of acid present based on the reaction's stoichiometry. This principle is fundamental in solving titration problems efficiently, as seen in the given exercise.

Stoichiometry

Stoichiometry is a key element of chemistry that quantifies the relationships between reactants and products in chemical reactions. It involves using balanced chemical equations to calculate the moles of substances involved.
In the context of a titration involving a weak monoprotic acid and a strong base like NaOH, stoichiometry helps us determine how the moles of acid correlate with moles of the base. In our example, noticing the one-to-one correspondence: one mole of acid neutralizes one mole of base.
Understanding stoichiometry allows chemists to relate quantities like volume and concentration to the number of moles involved, which is vital for calculating unknowns in reactions, especially in titration analysis. Proper stoichiometry ensures that we can precisely calculate the original concentration of a solution based on its reaction with a known solution.