Problem 44 - Chemistry: The Central Science

Question

\mathrm{~A} 20.0\( - \)\mathrm{mL}\( sample of \)0.150 \mathrm{M} \mathrm{KOH}\( is titrated with \)0.125 \mathrm{M}\( \)\mathrm{HClO}_{4}\( solution. Calculate the \)\mathrm{pH}\( after the following volumes of acid have been added. (a) \)20.0 \mathrm{~mL}\(, (b) \)23.0 \mathrm{~mL}\(, (c) \)24.0 \mathrm{~mL}\(. (d) \)25.0 \mathrm{~mL}\( (e) \)30.0 \mathrm{~mL}$

Step-by-Step Solution

Answer

The pH values after adding the specified volumes of HClO₄ are: (a) 12.10 (b) 11.46 (c) 7.00 (d) 2.59 (e) 1.78

Step 1: Write the reaction equation

The balanced reaction between KOH (a strong base) and HClO4 (a strong acid) is: \(KOH(aq) + HClO_4 (aq) \rightarrow KClO_4 (aq) + H_2O(l)\) Since both KOH and HClO4 are strong, they will completely dissociate in water and react with each other in a 1:1 ratio. It is essential to recognize this stoichiometric relationship.

Step 2: Find initial moles of the base

We start with 20.0 mL of 0.150 M KOH. To calculate moles, use the formula: moles = volume × molarity moles of KOH = (20.0 mL)(0.150 mol/L) = 3.00 mmol

Step 3: Calculate moles of the acid added for each volume

For each volume, calculate the moles of HClO4 added: (a) 20.0 mL: (20.0 mL)(0.125 mol/L) = 2.50 mmol (b) 23.0 mL: (23.0 mL)(0.125 mol/L) = 2.88 mmol (c) 24.0 mL: (24.0 mL)(0.125 mol/L) = 3.00 mmol (d) 25.0 mL: (25.0 mL)(0.125 mol/L) = 3.13 mmol (e) 30.0 mL: (30.0 mL)(0.125 mol/L) = 3.75 mmol

Step 4: Find the leftover moles of base (OH⁻) or acid (H₃O⁺) for each volume

Note: If moles of KOH > moles of HClO4, there will be leftover OH⁻. If moles of KOH < moles of HClO4, there will be leftover H₃O⁺. (a) 2.50 mmol of HClO4 is added: 3.00 mmol - 2.50 mmol = 0.50 mmol OH⁻ left (b) 2.88 mmol of HClO4 is added: 3.00 mmol - 2.88 mmol = 0.12 mmol OH⁻ left (c) 3.00 mmol of HClO4 is added: Neutralization occurs, no leftover ions (d) 3.13 mmol of HClO4 is added: 3.13 mmol - 3.00 mmol = 0.13 mmol H₃O⁺ left (e) 3.75 mmol of HClO4 is added: 3.75 mmol - 3.00 mmol = 0.75 mmol H₃O⁺ left

Step 5: Calculate pH for each volume

Using the moles of leftover OH⁻ or H₃O⁺, calculate the concentrations then find the pH or pOH as needed. Remember that \(pH + pOH = 14\). (a) [OH⁻]= (0.50 mmol) / (20.0 mL + 20.0 mL) = 0.0125 M, pOH = -log(0.0125) = 1.90, pH = 14 - 1.90 = 12.10 (b) [OH⁻]= (0.12 mmol) / (20.0 mL + 23.0 mL) = 0.00286 M, pOH = -log(0.00286) = 2.54, pH = 14 - 2.54 = 11.46 (c) Neutralization point, pH = 7 (d) [H₃O⁺]= (0.13 mmol) / (20.0 mL + 25.0 mL) = 0.00260 M, pH = -log(0.00260) = 2.59 (e) [H₃O⁺]= (0.75 mmol) / (20.0 mL + 30.0 mL) = 0.0167 M, pH = -log(0.0167) = 1.78

Step 6: Final Answer

The pH values after adding the specified volumes of HClO₄ are: (a) 12.10 (b) 11.46 (c) 7.00 (d) 2.59 (e) 1.78

Key Concepts

pH Calculation

Understanding the calculation of pH is crucial when analyzing acid-base reactions. pH is a measure of the acidity or basicity of an aqueous solution, and it is calculated as the negative logarithm (base 10) of the concentration of hydrogen ions (\text{H}^{+}) or hydronium ions (\text{H}_3\text{O}^{+}) in the solution. The formula for pH is expressed as:
\[ pH = -\text{log}([H_3O^+]) \]
Similarly, for solutions with excess hydroxide ions (\text{OH}^{-}), we measure the pOH, which can be converted to pH using the relationship \(pH + pOH = 14\), assuming standard room temperature. The steps provided in the solution highlight how to calculate the pH or pOH based on the remaining amount of acid or base after a titration. Afterwards, if the OH- concentration is available, as in parts (a) and (b) of the solution, we find pOH and then use the relationship to get the pH. Conversely, when H3O+ concentration is available, as in parts (d) and (e), we can directly calculate the pH.

Stoichiometry

Stoichiometry is the aspect of chemistry that deals with calculating the amounts of reactants and products in chemical reactions. In an acid-base titration such as the one presented in the problem, stoichiometry helps to determine the equivalent point of reaction where the number of moles of acid equals the number of moles of base. The given stoichiometry in this reaction is 1:1, meaning that one mole of KOH reacts with one mole of HClO4 to produce one mole of the salt KClO4 and one mole of water.
For each part of the exercise, the number of moles of the reactants (KOH and HClO4) was calculated by multiplying the volume of the solution used in the titration by the molarity, which reflects the concepts of stoichiometry. By utilizing these calculations, we can track the reaction’s progress. Specifically, we determine whether the reaction mixture is acidic, basic, or neutral at different stages of the titration.

Neutralization Reaction

A neutralization reaction occurs when an acid and a base react to form water and a salt, typically resulting in a change in pH toward neutrality. In the provided exercise, KOH (a strong base) reacts with HClO4 (a strong acid) in a neutralization reaction. When exactly equal moles of KOH and HClO4 are combined, such as in part (c), they neutralize each other completely, and the pH of the solution becomes 7, which is neutral for standard room temperature conditions.
It's important to realize that the point of neutralization is characterized by the absence of excess hydronium or hydroxide ions in the solution. In this example, before and after the neutralization point, the pH is dictated by the excess ions left over. Prior to reaching the neutralization point, excess hydroxide ions make the solution basic (pH > 7), while after the point, excess hydronium ions make it acidic (pH < 7). This concept is key in titration procedures to identify the end point of the titration, which is often done using pH indicators or pH meters in practical applications.