Question

A \(20.0-\mathrm{mL}\) sample of \(0.200 \mathrm{M} \mathrm{HBr}\) solution is titrated with \(0.200 \mathrm{M} \mathrm{NaOH}\) solution. Calculate the pH of the solution after the following volumes of base have been added: (a) \(15.0 \mathrm{~mL}\). (b) \(19.9 \mathrm{~mL}\) (c) \(20.0 \mathrm{~mL}\) (d) \(20.1 \mathrm{~mL}\), (e) \(35.0 \mathrm{~mL}\)

Short answer

For each of the scenarios: (a) After adding 15.0 mL of NaOH, we have: Moles of HBr = 4.00 mmol; Moles of NaOH = 3.00 mmol. The solution is acidic: [H+] = 0.05 M. Therefore, the pH is \(2\). (b) After adding 19.9 mL of NaOH, we have: Moles of HBr = 4.00 mmol; Moles of NaOH = 3.98 mmol. The solution is acidic: [H+] = 0.001 M. Therefore, the pH is \(3\). (c) After adding 20.0 mL of NaOH, we have: Moles of HBr = 4.00 mmol; Moles of NaOH = 4.00 mmol. The solution is neutral, and the pH is \(7\). (d) After adding 20.1 mL of NaOH, we have: Moles of HBr = 4.00 mmol; Moles of NaOH = 4.02 mmol. The solution is basic: [OH-] = 0.001 M. Therefore, the pH is \(11\). (e) After adding 35.0 mL of NaOH, we have: Moles of HBr = 4.00 mmol; Moles of NaOH = 7.00 mmol. The solution is basic: [OH-] = 0.05 M. Therefore, the pH is \(12\).

Step-by-step solution

Calculate the initial mmoles of HBr and NaOH

The initial mmoles of HBr and NaOH can be calculated by using their initial volumes and concentrations: Moles of HBr = Initial_volume_(HBr) × Initial_concentration_(HBr) Moles of NaOH = Added_volume_(NaOH) × Initial_concentration_(NaOH) For each situation (a, b, c, d, e), the moles of NaOH will change according to the given volume added, while the moles of HBr will stay the same, as the initial volume and concentration are constant.

Determine the mmoles of species after reaction

After adding NaOH to HBr, we'll have a reaction between these species to form water and a salt (NaBr). We can represent this reaction as: HBr + NaOH ⟶ H2O + NaBr We need to determine how many mmoles of HBr and NaOH are left after the reaction has occurred. To do this, we need to subtract the mmoles of the limiting reactant (the one that runs out first) from the mmoles of each species.

Calculate the pH

Depending on whether the solution is acidic, neutral, or basic after the reaction, we'll use different approaches to calculate the pH. 1. If there are mmoles of H+ ions left in the solution (acidic solution), the concentration of H+ ions can be calculated as: [H+] = Moles_of_HBr_left / Total_volume_of_solution Then, the pH can be calculated from the [H+] concentration using the formula: pH = -log([H+]) 2. If the solution is neutral, the pH will be 7. 3. If there are mmoles of OH- ions left in the solution (basic solution), the concentration of OH- ions can be calculated as: [OH-] = Moles_of_NaOH_left / Total_volume_of_solution Now, we calculate the pOH from the [OH-] concentration using the formula: pOH = -log([OH-]) Finally, we find the pH using the relationship: pH = 14 - pOH Repeat this step for each scenario (a, b, c, d, e) to find the pH of the solution after adding the specified volume of NaOH.

Key concepts

Molarity Calculations

Molarity is a fundamental concept in chemistry that expresses the concentration of a solution. It is defined as the number of moles of solute divided by the volume of the solution in liters. For the given exercise, we use molarity to find out how many moles of hydrogen bromide (HBr) and sodium hydroxide (NaOH) are present in their respective solutions before they start reacting with each other. To calculate the moles of a compound, you use the formula:\[\text{Moles} = \text{Volume (L)} \times \text{Molarity (M)}\]This allows us to determine how much of each substance is available to take part in the titration reaction. For example, the initial moles of HBr are calculated using its volume (20.0 mL, which is 0.020 L) and its concentration (0.200 M), which yields:\[\text{Moles of HBr} = 0.020 \text{ L} \times 0.200 \text{ M} = 0.004 \text{ moles}\]Performing similar calculations for NaOH, while considering each volume is crucial for later determining the limiting reactant in the titration process.

Acid-Base Reaction

In a titration involving an acid and a base, such as HBr and NaOH, a neutralization reaction occurs. This specific reaction is categorized as an acid-base reaction where an acid (HBr) reacts with a base (NaOH) to produce water (H₂O) and a salt (NaBr).The chemical equation representing this reaction is:\[\text{HBr} + \text{NaOH} \rightarrow \text{H}_2\text{O} + \text{NaBr}\]This is a classic example of a one-to-one stoichiometric reaction. This means that one mole of HBr reacts with one mole of NaOH. In practical terms, this means when equal numbers of moles of the two react (i.e., they are completely used up), the resulting solution becomes neutral. If there's an excess of either HBr or NaOH after the reaction, the solution will be acidic or basic, respectively.

pH Determination

Determining the pH of a solution during a titration is essential for understanding its acidity or basicity. The pH scale ranges from 0 to 14, with lower numbers indicating acidic solutions, 7 being neutral, and higher numbers indicating basic solutions. In this exercise, for any scenario where HBr remains after titration, the solution will be acidic. We calculate the pH by first finding the concentration of hydrogen ions \([\text{H}^+]\) and then using the formula:\[pH = -\log([\text{H}^+])\]If NaOH is in excess, meaning some is left unreacted after neutralization, the solution is basic. Here, we first calculate the concentration of hydroxide ions \([\text{OH}^-]\), find the pOH:\[pOH = -\log([\text{OH}^-])\]and then determine pH using:\[pH = 14 - pOH\]Understanding how to interpret the pH based on the chemical reactions occurring gives valuable insight into the titration process.

Stoichiometry in Titrations

Stoichiometry allows us to predict how much of each substance will be involved in or produced by a chemical reaction. In titrations like the reaction between HBr and NaOH, stoichiometry helps determine the relative quantities of reactants and products. A key part of this process is identifying the limiting reactant – the substance that is completely consumed first, stopping the reaction. During the titration:

• Initial calculations determine the moles of HBr and added moles of NaOH.
• The moles of the limiting reactant are subtracted from both reactants to determine remaining amounts.
• The results guide us to calculate the solution's pH, as either excess HBr or NaOH will define the acidity or basicity.

This understanding of stoichiometry is crucial to accurately interpreting titrations, facilitating predictions about reaction outcomes and guiding the determination of solution characteristics post-reaction.