Problem 58 - Chemistry An Atoms First Approach

Question

Consider the titration of \(80.0 \mathrm{mL}\) of \(0.100 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) by \(0.400 M\) HCl. Calculate the \(p H\) of the resulting solution after the following volumes of HCl have been added. a. \(0.0 \mathrm{mL}\) b. \(20.0 \mathrm{mL}\) c. \(30.0 \mathrm{mL}\) d. \(40.0 \mathrm{mL}\) e. \(80.0 \mathrm{mL}\)

Step-by-Step Solution

Answer

a) \(pH \approx 13.301\) b) \(pH \approx 12.602\) Follow the same process for the remaining volumes to find the pH values for 30.0 mL, 40.0 mL, and 80.0 mL of HCl added.

Step 1: We must first find the initial amount of \(\mathrm{Ba(OH)_2}\) in moles. To do this, multiply the initial volume by the molarity: $$moles = 80.0\ \mathrm{mL} \times 0.100\ \mathrm{M} = 0.00800\ \mathrm{mol}.$$ #Step 2: Determine moles and concentrations of reagent ions at each point of the titration# For each of the given HCl volumes, calculate the moles of HCl and find the prevailing concentrations of \(\mathrm{OH^-}\) ions in the solution by accounting for the stoichiometry. If all of the \(\mathrm{OH^-}\) ions are consumed, calculate the moles of \(\mathrm{H_3O^+}\) ions generated. #Step 3: Calculate the pH at each stage of the titration# Use the concentrations of \(\mathrm{OH^-}\) and \(\mathrm{H_3O^+}\) ions to calculate the pH at every stage of the titration. Use the formula \(pH = 14 - pOH\) if the \(\mathrm{OH^-}\) ions haven't been consumed entirely and \(pH = -\log_{10} [\mathrm{H_3O^+}]\) if you need to find the pH using the \(\mathrm{H_3O^+}\) concentration. Now let's proceed to the specific calculation of the pH for each situation here: a) 0.0 \(\mathrm{mL}\) HCl added #Step 1a: Determine the Initial Moles of \(\mathrm{OH^-}\) ions#

Calculate the moles of \(\mathrm{OH^-}\) ions from the initial moles of \(\mathrm{Ba(OH)_2}\). As each mole of \(\mathrm{Ba(OH)_2}\) dissociates into two moles of \(\mathrm{OH^-}\) ions, we get: $$0.00800\ \mathrm{mol}\ \mathrm{Ba(OH)_2} \times 2\ \mathrm{mol}\ \mathrm{OH^-}/\mathrm{mol}\ \mathrm{Ba(OH)_2}=0.0160\ \mathrm{mol}\ \mathrm{OH^-}.$$ The concentration of \(\mathrm{OH^-}\) ions is: $$\frac{0.0160\ \mathrm{mol}\ \mathrm{OH^-}}{80.0\ \mathrm{mL}} = 0.200\ \mathrm{M}.$$ #Step 2a: Calculate \(pOH\) and \(pH\)#

Step 2: Since the concentration of \(\mathrm{OH^-}\) ions is known, we can find the \(pOH\): $$pOH = -\log_{10} [\mathrm{OH^-}] = -\log_{10} (0.200) = 0.699.$$ Now, we can find the pH: $$pH = 14 - 0.699 = 13.301.$$ b) \(20.0 \mathrm{mL}\) HCl added #Step 1b: Calculate the moles of \(\mathrm{OH^-}\) ions and HCl#

Before adding the HCl, we know there are \(0.0160\ \mathrm{mol}\) of \(\mathrm{OH^-}\) ions. To find how many moles of HCl are added, multiply the volume by the concentration: $$20.0\ \mathrm{mL} \times 0.400\ \mathrm{M} = 0.00800\ \mathrm{mol}.$$ #Step 2b: Calculate the moles and concentrations of leftover ions#

Step 3: In this case, two moles of HCl neutralize one mole of \(\mathrm{Ba(OH)_2}\). After the reaction, there will be \(0.00400\ \mathrm{mol}\) of \(\mathrm{OH^-}\) ions left in the solution: $$0.0160\ \mathrm{mol} - 2 \times (0.00800\ \mathrm{mol}) = 0.00400\ \mathrm{mol}.$$ The \(\mathrm{OH^-}\) concentration is: $$\frac{0.00400\ \mathrm{mol}}{80.0\ \mathrm{mL} + 20.0\ \mathrm{mL}} = 0.0400\ \mathrm{M}.$$ #Step 3b: Calculate \(pOH\) and \(pH\)#

Since the concentration of \(\mathrm{OH^-}\) ions is known, we can find the \(pOH\): $$pOH = -\log_{10} [\mathrm{OH^-}] = -\log_{10} (0.0400) = 1.398.$$ Now, we can find the pH: $$pH = 14 - 1.398 = 12.602.$$ To find the pH in the other situations, follow the same process for 30.0 mL, 40.0 mL, and 80.0 mL of HCl added.

Key Concepts

pH calculation

The pH of a solution is a measure of its acidity or alkalinity. It is calculated based on the concentration of hydrogen ions (\[ \text{H}^+ \] ) or hydronium ions (\[ \text{H}_3\text{O}^+ \] ). The formula to calculate pH is:

• For acidic solutions: \(\text{pH} = -\log_{10} \, [\textrm{H}_3\textrm{O}^+]\).
• For basic solutions, often given in terms of hydroxide ions (\[ \text{OH}^- \] ), we use the relation between pH and pOH: \( \text{pH} = 14 - \text{pOH} \)
• The pOH can be calculated as: \(\text{pOH} = -\log_{10} \, [\text{OH}^-]\)

It's crucial to determine the correct ion concentration, either \[ \text{H}_3\text{O}^+ \] for acidic solutions or \[ \text{OH}^- \] for basic solutions, to effectively calculate the pH. When the solution is neutral, pH is 7, which corresponds to pure water. Calculating the pH in titrations is central because it helps us understand how the solution's acidity changes as agents like acids or bases are added.

stoichiometry

Stoichiometry is the study of the relative quantities of reactants and products in chemical reactions. It involves using balanced chemical equations to calculate the amounts of substances consumed and produced.
Titration, being a chemical process, relies heavily on stoichiometry to determine how much of a compound is needed to react completely with another compound.

• Firstly, identify the balanced chemical equation of the reaction, as stoichiometric ratios are derived from these equations.
• It helps us predict the number of moles of one substance that will react with a given number of moles of another substance. This is particularly important in titration calculations.

In the specific reaction between \( \text{Ba(OH)}_2 \) and \( \text{HCl} \), stoichiometry tells us that one mole of \( \text{Ba(OH)}_2 \) reacts with two moles of \( \text{HCl} \). This ratio is essential for calculating how much \( \text{HCl} \) is needed to neutralize the \( \text{Ba(OH)}_2 \) present.

Ba(OH)2 and HCl reaction

The reaction between barium hydroxide ( \( \text{Ba(OH)}_2 \) ) and hydrochloric acid ( \( \text{HCl} \) ) is a classic acid-base reaction. This reaction is important in titration processes for its simplicity and clear demonstration of a neutralization reaction.
The balanced equation for this reaction is:

• \[ \text{Ba(OH)}_2 + 2 \text{HCl} \rightarrow \text{BaCl}_2 + 2 \text{H}_2\text{O} \]

Here, \( \text{Ba(OH)}_2 \) is the base and \( \text{HCl} \) is the acid. In this process:

• Each mole of \( \text{Ba(OH)}_2 \) provides two hydroxide ions ( \( \text{OH}^- \) ),
• While each mole of \( \text{HCl} \) provides one hydronium ion ( \( \text{H}_3\text{O}^+ \) ).

The \( \text{H}_3\text{O}^+ \) ions from \( \text{HCl} \) combine with the \( \text{OH}^- \) ions from \( \text{Ba(OH)}_2 \) to form water, greatly reducing the \( \text{OH}^- \) concentration and affecting the pH as a result. Understanding this reaction mechanism helps us predict the changes in pH during the titration.

neutralization reaction

A neutralization reaction occurs when an acid and a base react to form water and a salt, achieving a neutral pH (around 7). These reactions are pivotal in titrations because they assist in determining unknown concentrations.
In the titration of \( \text{Ba(OH)}_2 \) with \( \text{HCl} \), the neutralization is the endpoint we aim for. With this reaction:

• Hydroxide ions ( \( \text{OH}^- \) ) from \( \text{Ba(OH)}_2 \) neutralize hydronium ions ( \( \text{H}_3\text{O}^+ \) ) from \( \text{HCl} \).
• The products gained are water ( \( \text{H}_2\text{O} \) ) and barium chloride ( \( \text{BaCl}_2 \) ).

During the titration, as more \( \text{HCl} \) is added, \( \text{OH}^- \) are gradually consumed. The point where all \( \text{OH}^- \) ions are neutralized is called the equivalence point, marked by a significant change in pH. Knowing when the equivalence point is reached allows us to calculate the exact concentrations of the unknown solution, which is the goal of a titration experiment.