Question

Consider the titration of \(50.0 \mathrm{~mL}\) of \(0.116 \mathrm{M} \mathrm{NaOH}\) with \(0.0750\) M HCl. Calculate the \(\mathrm{pH}\) after the addition of each of the following volumes of acid: (a) \(5.0 \mathrm{~mL}\) (b) \(50 \mathrm{~mL}\) (c) \(0.10 \mathrm{~L}\)

Short answer

The pH values are approximately 12.99, 12.31, and 1.95 after 5.0 mL, 50 mL, and 100 mL of HCl are added, respectively.

Step-by-step solution

Initial Information and Reaction Setup

The reaction between NaOH and HCl is a strong base-strong acid titration, where the reaction can be represented as \( \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \). Start with 50.0 mL of 0.116 M NaOH. The number of moles of NaOH is calculated by \( 0.0500 \, \text{L} \times 0.116 \, \text{mol/L} = 0.0058 \, \text{mol} \). The concentration of HCl is 0.0750 M.

Calculate Moles of HCl Added

For part (a), calculate the moles of HCl added: \( 0.0050 \, \text{L} \times 0.0750 \, \text{mol/L} = 0.000375 \, \text{mol} \).For part (b), moles of HCl: \( 0.050 \, \text{L} \times 0.0750 \, \text{mol/L} = 0.00375 \, \text{mol} \).For part (c), moles of HCl: \( 0.100 \, \text{L} \times 0.0750 \, \text{mol/L} = 0.0075 \, \text{mol} \).

Calculate Remaining NaOH or Resulting pH

For part (a):- Initial moles of NaOH - moles of HCl added = remaining NaOH = \( 0.0058 - 0.000375 = 0.005425 \, \text{mol} \).- Volume after 5 mL HCl is \( 55 \, \text{mL} \), so concentration of NaOH = \( \frac{0.005425}{0.055} = 0.0987 \, \text{M} \).- \( \text{pH} = 14 - \text{pOH} = 14 + \log(0.0987) \approx 12.99 \).For part (b):- Remaining NaOH = \( 0.0058 - 0.00375 = 0.00205 \, \text{mol} \).- Volume after 50 mL HCl is \( 100 \, \text{mL} \), concentration = \( \frac{0.00205}{0.100} = 0.0205 \, \text{M} \).- \( \text{pH} = 14 + \log(0.0205) \approx 12.31 \).For part (c):- Excess HCl = \( 0.0075 - 0.0058 = 0.0017 \, \text{mol} \).- Total volume is \( 150 \, \text{mL} \), concentration = \( \frac{0.0017}{0.150} = 0.0113 \, \text{M} \).- \( \text{pH} = -\log(0.0113) \approx 1.95 \).

Determine Exact pH Values

For each scenario, the concentration of remaining NaOH or excess HCl helps determine the exact pH using pH and pOH calculations based on the formulae \( \text{pH} = 14 + \log[\text{OH}^-] \) for a basic solution and \( \text{pH} = -\log[\text{H}^+] \) for an acidic solution.

Key concepts

Strong Acid-Strong Base Titration

Titration is a laboratory technique used to determine the concentration of a solution through its reaction with a solution of known concentration. A strong acid-strong base titration involves a neutralization reaction where a strong acid like HCl reacts with a strong base such as NaOH, producing water and a salt (NaCl in this case).
This type of titration typically reaches an equivalence point where the amount of acid equals the amount of base present. Such reactions usually lead to a significant pH change at the equivalence point, which can be sharp and easily detectable using indicators or pH meters.

• At the start of the reaction, the solution's pH is determined by the strong base present.
• As the acidic titrant is added, it neutralizes the base, causing the pH to decrease.
• The pH continues to drop sharply until the equivalence point, where it becomes neutral.
• Beyond the equivalence point, excess acid lowers the pH further, leading to an acidic solution.

Understanding the nature of strong acid-strong base titrations allows us to predict these changes in pH during the titration process, which are essential for calculating concentrations and determining the pH values at different stages of the titration.

pH Calculation

The pH is a measure of the hydrogen ion concentration in a solution. For strong acid-strong base titrations, calculating the pH is critical at various stages for insight into the solution's acidity or basicity. The pH scale ranges from 0 to 14, where pH 7 is neutral.
During titration:

• The starting pH is often high if you begin with a strong base like NaOH, likely around pH 13-14.
• As HCl (a strong acid) is added, it begins to neutralize the NaOH, reducing the pH.
• Before reaching the equivalence point, the pH is determined by the unreacted OH⁻ ions. You can calculate it using the formula: \(\text{pH} = 14 + \log{[\text{OH}^-]}\).
• If the titration passes the equivalence point, excess H⁺ determines the pH, calculated as \(\text{pH} = -\log{[\text{H}^+]}\).

These calculations make it possible to find the exact pH after each amount of acid is added, crucial for thorough understanding and analysis of the titration curve.

Molarity

Molarity, denoted by M, is the concentration of a solution expressed as moles of solute per liter of solution. Molarity is extensively used in titration calculations because it allows for the precise measurement of reagent quantities needed to neutralize a given amount of solute.
In the context of titration, the molarity of the solutions involved plays a crucial role:

• The molarity of NaOH (the strong base) is given as 0.116 M. This helps in determining the initial moles of NaOH available for reaction.
• The molarity of HCl (the strong acid) is 0.0750 M, which helps calculate how many moles of acid are added at each step of the titration.
• Accurate molarity values are essential for stoichiometric calculations to ensure precise equivalence points are reached during titration experiments.

Knowing the initial molarity and using it to calculate the moles involved aids students in understanding the direct relationship between molarity, volume, and moles—a foundational concept in chemistry.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions, based on the principle of conservation of mass. For titrations, stoichiometry dictates the proportions in which reactants react.
In a strong acid-strong base titration:

• The reaction equation is NaOH + HCl → NaCl + H₂O, which is balanced, indicating a 1:1 mole ratio between the reactants.
• Using stoichiometry, the moles of base initially present can be calculated using its volume and molarity.
• For each addition of acid, the moles of acid react completely with the moles of base, permitting calculation of any remaining reactant.
• These calculations help determine whether NaOH is in excess, if the solution is neutral at the equivalence point, or if excess HCl is present.

Through stoichiometry, students can calculate how far a reaction proceeds and predict changes in pH as seen in titration, reinforcing their understanding of quantitative aspects of chemical reactions.