Question

A 20.0-mL sample of \(0.200 \mathrm{M}\) HBr solution is titrated with \(0.200 \mathrm{M} \mathrm{NaOH}\) solution. Calculate the \(\mathrm{pH}\) 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

The pH values of the solution after the given volumes of NaOH have been added are as follows: (a) 15.0 mL: pH = 1.60 (b) 19.9 mL: pH = 3.30 (c) 20.0 mL: pH = 7.00 (d) 20.1 mL: pH = 10.70 (e) 35.0 mL: pH = 12.70

Step-by-step solution

Determine the initial moles of HBr

To find the initial moles of HBr, we use the formula: Moles = Molarity × Volume where the molarity of HBr is 0.200 M and the volume is 20.0 mL (0.020 L). Moles of HBr = 0.200 M × 0.020 L = 0.004 mol

Calculate the moles of NaOH added for each case

We will use the same formula as in step 1 to calculate the moles of NaOH added in each case. Using molarity of NaOH = 0.200 M and given volumes: (a) Moles of NaOH added = 0.200 M × 0.015 L = 0.003 mol (b) Moles of NaOH added = 0.200 M × 0.0199 L = 0.00398 mol (c) Moles of NaOH added = 0.200 M × 0.020 L = 0.004 mol (d) Moles of NaOH added = 0.200 M × 0.0201 L = 0.00402 mol (e) Moles of NaOH added = 0.200 M × 0.035 L = 0.007 mol

Determine the moles of HBr and NaOH left after the reaction

For strong acid-strong base titrations, each mole of acid reacts with one mole of base, the moles left to calculate new equilibrium are as follows: (a) 0.004 mol HBr - 0.003 mol NaOH = 0.001 mol HBr left, no NaOH left (b) 0.004 mol HBr - 0.00398 mol NaOH = 0.00002 mol HBr left, no NaOH left (c) 0.004 mol HBr - 0.004 mol NaOH = no HBr left, no NaOH left (equivalence point) (d) no HBr left, 0.00402 mol NaOH - 0.004 mol HBr = 0.00002 mol NaOH left (e) no HBr left, 0.007 mol NaOH - 0.004 mol HBr = 0.003 mol NaOH left

Calculate the pH of the solution for each case

(a) Concentration of HBr left = 0.001 mol / (0.020 L + 0.015 L) = 0.025 M; pH = -log(0.025) = 1.60 (b) Concentration of HBr left = 0.00002 mol / (0.020 L + 0.0199 L) = 0.0005 M; pH = -log(0.0005) = 3.30 (c) At the equivalence point, the pH is neutral, so pH = 7.00 (d) Concentration of NaOH left = 0.00002 mol / (0.020 L + 0.0201 L) = 0.0005 M; pOH = -log(0.0005) = 3.30; pH = 14 - 3.30 = 10.70 (e) Concentration of NaOH left = 0.003 mol / (0.020 L + 0.035 L) = 0.05 M; pOH = -log(0.05) = 1.30; pH = 14 - 1.30 = 12.70 Now we have the pH values for all cases: (a) pH = 1.60 (b) pH = 3.30 (c) pH = 7.00 (d) pH = 10.70 (e) pH = 12.70

Key concepts

Acid-Base Reaction

Titration involves a fascinating chemical process called the acid-base reaction. Here, an acid reacts with a base to produce water and a salt. It's important to note that this type of reaction is usually very fast and goes to completion. Both the acid and the base dissociate completely in water to provide Hydrogen ions (\( H^+ \)) from the acid and Hydroxide ions (\( OH^- \)) from the base. This results in the production of water \( H_2O \).

In our exercise, hydrobromic acid (HBr) serves as the acid, and sodium hydroxide (NaOH) is the base. The overall reaction is: \[ \text{HBr} + \text{NaOH} \rightarrow \text{H}_2\text{O} + \text{NaBr} \]
The reaction's one-to-one stoichiometry means each mole of HBr reacts with one mole of NaOH. This makes it straightforward to determine what is left in the solution after the reaction takes place.

pH Calculation

Calculating pH is a pivotal part of understanding titration and acid-base reactions. pH is a measure of the hydrogen ion concentration in a solution and is given by the formula: \[ \text{pH} = -\log [H^+] \]
In the exercise, the pH varies as different amounts of NaOH are added. It is essential to track the moles of HBr and NaOH at each stage:

• Before the equivalence point, the solution is acidic because all NaOH is consumed with some HBr still remaining, which affects the pH calculations to be below 7.
• At the equivalence point, all HBr is neutralized, leading to a neutral pH of 7.
• Beyond the equivalence point, there are leftover NaOH moles, making the solution basic, hence a pH above 7.

Understanding how to calculate pH during titration helps in observing how the solution's nature changes from acidic to neutral to basic based on the points of reaction.

Equivalence Point

The equivalence point is a critical concept in titration. It is the stage where amounts of acid and base solutions added are just sufficient to completely neutralize each other. Here, the number of moles of the acid initially present is equal to the number of moles of the base added.
In the example, the equivalence point is achieved when 20.0 mL of 0.200 M NaOH is added to 20.0 mL of 0.200 M HBr. At this point, the pH is neutral, typically around 7 for strong acid-strong base titrations. It's important to note that at the equivalence point:

• The solution doesn't have surplus HBr or NaOH, hence the neutral pH.
• Represents a key stage for determining the concentration of unknown solutions in a titration analysis.

Stoichiometry

Stoichiometry is the backbone of calculating titrations. It's about using the balanced chemical equation to determine the proportions of reactants and products involved.
In this exercise, the stoichiometric ratio of HBr to NaOH is 1:1, which means for every mole of HBr, one mole of NaOH reacts. This allows a straightforward calculation of moles remaining at different titration points.

• The volumes and molarity help calculate how many moles are initially present using \( \text{Moles} = \text{Molarity} \times \text{Volume} \).
• Adjust calculations for either acid or base remaining after the reaction to find the new concentration for subsequent pH calculations.

Understanding stoichiometry allows us to predict the outcome of mixing chemicals and is essential for achieving desired concentrations in titration and other chemical reactions.