Question

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

Short answer

The pH at each point of titration is: (a) 13.09, (b) 12.52, (c) 7.00, (d) 4.48, and (e) 3.93.

Step-by-step solution

Write the reaction equation between the base and the acid

The reaction between the strong base KOH and the strong acid HClO₄ can be represented by the following balanced chemical equation: \(KOH + HClO_4 \rightarrow KClO_4 + H_2O\)

Calculate the initial moles of KOH

We are given a 20.0 mL sample of 0.150 M KOH. To find the initial moles of KOH, use the formula: moles = Molarity × Volume in liters moles of KOH = 0.150 M × 0.020 L = 0.003 mol

Calculate the moles of HClO₄ added for each volume

The given concentration of HClO₄ is 0.125 M. We will calculate the moles of HClO₄ for the respective added volumes. (a) 20.0 mL: moles of HClO₄ = 0.125 M × 0.020 L = 0.00250 mol (b) 23.0 mL: moles of HClO₄ = 0.125 M × 0.023 L = 0.002875 mol (c) 24.0 mL: moles of HClO₄ = 0.125 M × 0.024 L = 0.00300 mol (d) 25.0 mL: moles of HClO₄ = 0.125 M × 0.025 L = 0.003125 mol (e) 30.0 mL: moles of HClO₄ = 0.125 M × 0.030 L = 0.00375 mol

Determine the pH at each point of titration

We know the relation between moles of KOH and moles of HClO₄, pH can be calculated as follows: (a) 0.003 mol KOH - 0.00250 mol HClO₄ = 0.0005 mol KOH left pH = 14 - pOH pOH = -log(OH⁻ molarity) pH = 14 - pOH(KOH) = 14 - (-log((0.0005 mol)/(0.040 L))) = 13.09 (b) 0.003 mol KOH - 0.002875 mol HClO₄ = 0.000125 mol KOH left pH = 14 - pOH(KOH) = 14 - (-log((0.000125 mol)/(0.043 L))) = 12.52 (c) 0.003000 mol KOH - 0.003000 mol HClO₄ = 0 (neutralization point) pH = 7 (d) 0.003000 mol KOH - 0.003125 mol HClO₄ = 0.000125 mol HClO₄ left (excess acid) pH = -log(H⁺ molarity) pH = -log((0.000125 mol)/(0.045 L)) = 4.48 (e) 0.003000 mol KOH - 0.003750 mol HClO₄ = 0.00075 mol HClO₄ left (excess acid) pH = -log((0.00075 mol)/(0.050 L)) = 3.93 The pH at each point of titration is: (a) 13.09 (b) 12.52 (c) 7.00 (d) 4.48 (e) 3.93

Key concepts

Acid-Base Reactions

Acid-base reactions are fundamental chemical reactions in which an acid reacts with a base to produce water and a salt. They are a key aspect of titration experiments. In our case, we have a reaction between a strong base, potassium hydroxide (KOH), and a strong acid, perchloric acid (\(\text{HClO}_4\)). This reaction can be represented in a simple balanced chemical equation as follows:
\[\text{KOH} + \text{HClO}_4 \rightarrow \text{KClO}_4 + \text{H}_2\text{O}\]

• The strong base dissociates completely in water to produce hydroxide ions (OH⁻).
• The strong acid dissociates completely in water to produce hydrogen ions (\(\text{H}^+\)).
• In the reaction, the hydrogen ions from the acid combine with the hydroxide ions from the base to form water, a neutral compound.

These reactions are crucial in determining various characteristics of a substance, such as pH levels, and are useful in applications like determining the concentration of unknown solutions.

pH Calculation

Calculating the pH of a solution during a titration involves understanding the concentration of hydrogen ions in the solution. pH is a measure of how acidic or basic a solution is, with lower pH values being more acidic, higher values being more basic, and 7 being neutral.
The formula used to calculate pH is:
\[-\log[\text{H}^+]\]
For bases, we often calculate the pOH first, since bases release hydroxide ions (OH⁻) and not hydrogen ions directly. The pH can then be found using:
\[\text{pH} = 14 - \text{pOH}\]

• For a strong base like KOH, dissociate to give OH⁻, calculate the pOH using: \[\text{pOH} = -\log(\text{[OH⁻]})\]
• The remaining concentration of OH⁻ determines how basic the solution is after reacting a specific amount of acid.
• For strong acids like HClO⁴, calculate pH directly from the concentration of remaining H⁺ ions after the base has been neutralized.

By knowing the amounts of acid and base added, and any excess left, you can compute the exact pH at various stages of the titration.

Neutralization Point

The neutralization point in a titration occurs when the amount of acid equals the amount of base, resulting in the formation of water and a salt, leaving no excess acid or base in the solution. In such reactions between strong acids and strong bases, this is known as the equivalence point, and it is where the pH is typically neutral, around 7.00.
In the given titration exercise:

• Both KOH and HClO₄ are completely dissociated in water, making it a perfect scenario for reaching a true neutralization point.
• The equivalence point is reached when 24.0 mL of 0.125 M HClO₄ completely reacts with 20.0 mL of 0.150 M KOH.
• At this point, the moles of H⁺ from the acid equal the moles of OH⁻ from the base, resulting in pure water and a neutral pH of 7.00.

Reaching the neutralization point is crucial for accurate chemical analysis, as it indicates the amount of substance needed to neutralize the other.

Strong Acid and Strong Base Reaction

When strong acids and strong bases react, they undergo complete ionization, meaning they fully dissociate in water and react completely with each other to form water and a salt. This results in very distinct pH changes that are sharp and predictable.
Any added amount of the strong acid will be neutralized by an equivalent amount of the strong base, until one is in excess.

• Before reaching the equivalence point, any added acid or base will lead to an increase in either H⁺ or OH⁻ concentration, depending on which is in excess, affecting the pH values significantly.
• If the base (like KOH) is in excess, the solution remains basic with a high pH.
• Once the equivalence point is surpassed, any additional acid results in the solution becoming acidic, reducing the pH rapidly.

Such titrations provide a clear indication of pH changes at different stages, making them highly practical for determining concentration and thorough understanding of acid-base chemistry.