Problem 44 - Chemistry The Central Science In Si Units

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 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

(a) pH = 7.0, (b) pH ≈ 2.06, (c) pH ≈ 6.5, (d) pH ≈ 2.56, (e) pH ≈ 1.82.

Step 1: Initial Calculation of Moles

First, calculate the moles of KOH present in the 20.0 mL sample. The concentration of KOH is given as 0.150 M. \[ \text{Moles of } \mathrm{KOH} = 0.150 \times \frac{20.0}{1000} = 0.003 \text{ moles}. \]

Step 2: Determine Moles of HClO4 for Each Volume

Now calculate the moles of \(\mathrm{HClO}_4\) added for each volume given:- For 20.0 mL: \[ \text{Moles of } \mathrm{HClO}_4 = 0.125 \times \frac{20.0}{1000} = 0.0025 \text{ moles}. \]- For 23.0 mL: \[ \text{Moles of } \mathrm{HClO}_4 = 0.125 \times \frac{23.0}{1000} = 0.002875 \text{ moles}. \]- For 24.0 mL: \[ \text{Moles of } \mathrm{HClO}_4 = 0.125 \times \frac{24.0}{1000} = 0.003 \text{ moles}. \]- For 25.0 mL: \[ \text{Moles of } \mathrm{HClO}_4 = 0.125 \times \frac{25.0}{1000} = 0.003125 \text{ moles}. \]- For 30.0 mL: \[ \text{Moles of } \mathrm{HClO}_4 = 0.125 \times \frac{30.0}{1000} = 0.00375 \text{ moles}. \]

Step 3: Calculate pH at Volume 20.0 mL (Equivalence Point)

At this point, moles of \(\mathrm{HClO}_4\) = moles of \(\mathrm{KOH}\) = 0.0025. The reaction is neutralized, so pH = 7.0 in neutral solution.

Step 4: Calculate pH at Volume 23.0 mL (Excess Acid)

There is excess \(\mathrm{HClO}_4\) since 0.002875 moles \(\mathrm{HClO}_4\) - 0.0025 moles \(\mathrm{KOH}\) = 0.000375 moles excess \(\mathrm{HClO}_4\). Total solution volume = 20.0 mL + 23.0 mL = 43.0 mL. \[ \text{Concentration of } \mathrm{H^+} = \frac{0.000375}{0.043} = 0.00872 \text{ M} \] \[ \text{pH} = -\log(0.00872) \approx 2.06. \]

Step 5: Calculate pH at Volume 24.0 mL (Slightly Past Equivalence)

Slightly past equivalence, 0.003 moles \(\mathrm{HClO}_4\) - 0.003 moles \(\mathrm{KOH}\) means no excess acid. The solution is dissociated far past zero, pH still continues beyond neutrality, though drastically low.\[ \text{Total volume is 44.0 mL. Use simply} \; pH \approx 7, \, \text{but tips lower here to acidically less neutral. } \approx 6.5 \]

Step 6: Calculate pH at Volume 25.0 mL (Excess Acid)

There is significant excess \(\mathrm{HClO}_4\): 0.003125 - 0.003 = 0.000125 moles. Total solution volume: 45.0 mL. \[ \text{Concentration of } \mathrm{H^+} = \frac{0.000125}{0.045} = 0.00278 \text{ M} \] \[ \text{pH} = -\log(0.00278) \approx 2.56. \]

Step 7: Calculate pH at Volume 30.0 mL (More Excess Acid)

There is notable excess **HClO4**: **0.00375 moles - 0.003 moles = 0.00075 moles** excess. Total volume: **50.0 mL.** \[\text{Concentration of } \mathrm{H^+} = \frac{0.00075}{0.050} = 0.015 \text{ M} \]\[ \text{pH} = -\log(0.015) \approx 1.82. \]

Key Concepts

pH calculation

pH calculation is a fundamental concept in acid-base chemistry and is essential for understanding how acidic or basic a solution is. The pH value is calculated using the formula: - \( ext{pH} = - ext{log}_{10}[ ext{H}^+] \), where \([ ext{H}^+]\) is the concentration of hydrogen ions in the solution. For instance, if the concentration of \([ ext{H}^+]\) is 0.00278 M, the pH is calculated as:- \( ext{pH} = - ext{log}_{10}(0.00278) \approx 2.56 \). This indicates an acidic medium, as pH values below 7 are considered acidic. Conversely, a pH value above 7 indicates a basic solution. A neutral solution, like pure water, has a pH of 7. Calculations during titrations often show how pH changes as acids are added to bases or vice versa, allowing one to monitor the neutrality of the solution.

equivalence point

In a titration, the equivalence point is reached when the amount of acid equals the amount of base in the solution. This is a crucial stage, as it indicates that the chemical reaction is complete. Typically, this results in a neutral pH of around 7 for strong acid-strong base reactions, where both components fully disassociate into ions. For example, in a base-acid titration involving KOH and HClO4, the equivalence point occurs when 20 mL of the acid neutralizes the 20 mL of 0.150 M KOH. This results in both substances having equal moles, calculated to be 0.003 moles in the initial setup. - Neutralization at the equivalence point often shows a clear pH shift on the titration curve, which can be visualized as an abrupt change. Understanding the equivalence point helps to predict when a titration should stop and can indicate proper stoichiometric measurements in reactions.

molar concentration

Molar concentration, often called molarity, is a measure of the amount of a solute dissolved in a solution. It is denoted as M and measured in moles per liter. Understanding molarity is essential for performing pH calculations and titration analysis.- For example, if you have a solution of KOH with a concentration of 0.150 M and a volume of 20.0 mL, the moles of KOH can be calculated using the formula: \[ \text{Moles of } KOH = ext{Molarity} \times \text{Volume in Liters} = 0.150 \times \frac{20.0}{1000} = 0.003 \text{ moles}. \]During titrations, knowing the molar concentration helps to identify how much of the solution is needed to reach the equivalence point. This information proves crucial for accurate calculation of pH levels at different stages of the titration process, ensuring precise reactions occur in experimental setups.