Problem 44 - Chemistry The Central Science

Question

A 20.0 -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 \(\mathrm{mL}\) \((\mathbf{c}) 24.0 \mathrm{mL},(\mathbf{d}) 25.0 \mathrm{mL},(\mathbf{e}) 30.0 \mathrm{mL}\)

Step-by-Step Solution

Answer

The pH values of the solution at each point during the titration with HClO₄ are: a) 11.70 b) 11.30 c) 7.00 d) 2.52 e) 2.12

Step 1: Write the balanced chemical equation

The balanced chemical equation for the reaction between KOH and HClO4 is: \(KOH(aq) + HClO_4(aq) \rightarrow KClO_4(aq) + H_2O(l)\)

Step 2: Determine the moles of KOH and HClO4 added to the solution

For each volume of HClO4 (20.0 mL, 23.0 mL, 24.0 mL, 25.0 mL, and 30.0 mL), we will calculate the moles of KOH and HClO4: Initial moles of KOH in 20.0 mL solution = 0.150 M * 0.020 L = 0.0030 mol

Step 3: Calculate moles of HClO4 added at each point

For each given volume of HClO4, the moles of HClO4 added can be calculated using the given concentration: a) 20.0 mL of HClO4: (0.125 M) * (0.020 L) = 0.0025 mol b) 23.0 mL of HClO4: (0.125 M) * (0.023 L) = 0.002875 mol c) 24.0 mL of HClO4: (0.125 M) * (0.024 L) = 0.0030 mol d) 25.0 mL of HClO4: (0.125 M) * (0.025 L) = 0.003125 mol e) 30.0 mL of HClO4: (0.125 M) * (0.030 L) = 0.00375 mol

Step 4: Calculate moles of KOH and HClO4 at each point

We must take into account the moles reacting at each point, to decide if we must calculate the pH with the excess reactant, or the equilibrium of the solution: a) 20.0 mL of HClO4: Moles of \(KOH (0.0030 - 0.0025) = 0.0005\) mol, No leftover HClO4 b) 23.0 mL of HClO4: Moles of \(KOH (0.0030 - 0.002875) = 0.000125\) mol, No leftover HClO4 c) 24.0 mL of HClO4: At equivalence point, no leftover KOH or HClO4 d) 25.0 mL of HClO4: No leftover KOH, moles of \(HClO_4 (0.003125 - 0.0030) = 0.000125\) mol e) 30.0 mL of HClO4: No leftover KOH, moles of \(HClO_4 (0.00375 - 0.0030) = 0.00075\) mol

Step 5: Calculate the pH at each point

To calculate the pH at each point, we need to find the concentration of H3O+ or OH- and then use the pH or pOH formula. If there's excess KOH, calculate the pOH and then find the pH. If there's excess HClO4, calculate the pH directly. a) 20.0 mL of HClO4: Excess KOH, pOH = -log10([OH-]) = -log10(0.0005 mol / 0.040 L) = 2.30; pH = 14 - pOH = 11.70 b) 23.0 mL of HClO4: Excess KOH, pOH = -log10([OH-]) = -log10(0.000125 mol / 0.043 L) = 2.70; pH = 14 - pOH = 11.30 c) 24.0 mL of HClO4: Equivalence point, pH = 7 d) 25.0 mL of HClO4: Excess HClO4, pH = -log10([H3O+]) = -log10(0.000125 mol / 0.045 L) = 2.52 e) 30.0 mL of HClO4: Excess HClO4, pH = -log10([H3O+]) = -log10(0.00075 mol / 0.050 L) = 2.12 In conclusion, the pH values are as follows: a) 11.70 b) 11.30 c) 7.00 d) 2.52 e) 2.12

Key Concepts

pH calculation

When we talk about pH calculations, we're dealing with the acidity or basicity of a solution. The pH scale ranges from 0 to 14, where a lower number means higher acidity and a higher number indicates increased basicity. Neutral solutions have a pH of 7. To find the pH of a solution when you know the concentration of hydrogen ions \([H^{+}]\), use the formula: \( ext{pH} = - ext{log}_{10}([H^{+}])\).
For bases, such as KOH, you might first calculate the pOH using the hydroxide ion concentration \([OH^{-}]\), then convert it to pH using the relationship: \( ext{pH} + ext{pOH} = 14\). This conversion helps us find the pH for solutions with excess base, like in the initial parts of the titration problem where KOH was not completely neutralized.
By carefully tracking the changing concentrations during a titration, we can determine the varying pH at different stages in the process.

Neutralization reaction

Neutralization reactions occur when an acid and base react together to form water and a salt. This process is pivotal in the context of acid-base titrations. The balanced chemical equation involved in the given exercise is \(KOH + HClO_4 \rightarrow KClO_4 + H_2O\). These reactions typically result in the formation of water, which happens because of the combination of H\(^+\) from the acid and OH\(^-\) from the base.
Neutralization results in the balancing of acids and bases. When calculating the results of neutralization, it's important to ensure stoichiometric equivalence, where moles of H\(^+\) are equal to moles of OH\(^-\). Look at excess reactants to determine if the solution will be acidic or basic after the reaction completes. This will then help in determining the pH.

Equivalence point

The equivalence point in a titration is where the amount of acid equals the amount of base, leading to complete neutralization. At this stage, there are no excess reactants left in the solution.
For example, when 24 mL of \(HClO_4\) was added to the 20 mL of \(KOH\), the solution reached its equivalence point as 0.003 moles of both acid and base were present, leading the pH to be neutral at 7. This is because the presence of equal moles means all the acid and base has reacted completely, and none remains to affect the pH.
Detecting the equivalence point involves precise measurements and often the use of indicators or pH meters to observe the rapid shift in pH indicating neutralization.

Mole calculation

Mole calculations form the foundation of stoichiometry in chemistry and are crucial for titration exercises. Moles are a measure of quantity representing the number of atoms or molecules in a substance. Calculating moles accurately enables us to understand how much of each reactant is needed for complete reactions.
To find moles, use the formula \( ext{moles} = ext{concentration} \times ext{volume} \), where concentration is in molarity (M) and volume is in liters. For example, in the exercise, we calculated moles of \(KOH\) and \(HClO_4\) by using their given concentrations and volumes: \(0.15 \text{ M} \times 0.020 \text{ L} = 0.003 \text{ mol KOH}\).
Understanding moles supports the entire titration process, as it guides through the steps of finding out how different substances will react, reach the equivalence point, and determining pH changes throughout the procedure.