Question

A student is asked to determine the molarity of \(25.00 \mathrm{~mL}\) of a solution of \(\mathrm{HClO}_{4}\). He uses \(0.731 \mathrm{M} \mathrm{KOH}\). After adding \(42.35 \mathrm{~mL}\) of \(\mathrm{KOH}\), he realizes that he forgot to add an indicator. His TA suggests he take the \(\mathrm{pH}\) of the solution. The \(\mathrm{pH}\) is 12.39 (a) Did the student go beyond the equivalence point? (b) What is the molarity of the strong acid? (c) If he did, how many milliliters of \(\mathrm{KOH}\) did he add in excess? If he did not, how much more (in mL) KOH should he add?

Short answer

Answer: Yes, the student went past the equivalence point. The molarity of the strong acid (\(\mathrm{HClO}_{4}\)) is \(1.2368 \mathrm{M}\), and the student added \(5.71 \ \mathrm{mL}\) of \(\mathrm{KOH}\) in excess.

Step-by-step solution

Determine the moles of \(\mathrm{KOH}\) added

First, let's find the moles of \(\mathrm{KOH}\) added by the student. We can use the equation: moles = molarity \(\times\) volume. Moles of \(\mathrm{KOH}\) = \(0.731 \mathrm{M} \times 0.04235 \mathrm{L}\) = \(0.03092\) moles

Determine the moles of \(\mathrm{HClO}_{4}\) in the solution

Since \(\mathrm{HClO}_{4}\) and \(\mathrm{KOH}\) react in a 1:1 ratio, the moles of \(\mathrm{HClO}_{4}\) in the solution before the student added \(\mathrm{KOH}\) can be determined by the number of moles of \(\mathrm{KOH}\) added. Moles of \(\mathrm{HClO}_{4}\) = moles of \(\mathrm{KOH}\) Moles of \(\mathrm{HClO}_{4}\) = \(0.03092\) moles

Calculate the molarity of \(\mathrm{HClO}_{4}\)

Now, we can find out the molarity of the \(\mathrm{HClO}_{4}\) solution. We can use the equation: Molarity = moles/volume Molarity of \(\mathrm{HClO}_{4}\) = moles of \(\mathrm{HClO}_{4} / 0.02500 \mathrm{L}\) = \(1.2368 \mathrm{M}\) So, the molarity of the strong acid (\(\mathrm{HClO}_{4}\)) is \(1.2368 \mathrm{M}\).

Determine if the student went past the equivalence point

We know that strong acids and strong bases react to form a neutral solution, which has a pH of 7. If the pH after the reaction is greater than 7, the student went past the equivalence point. In this case, the pH is 12.39, which is greater than 7. Therefore, the student went beyond the equivalence point.

Calculate the excess \(\mathrm{KOH}\) added

Since the pH of the solution at the equivalence point would be 7, we can use the relation between \(\mathrm{OH}^{-}\) ion concentration and pH to find out the excess amount of \(\mathrm{KOH}\) added. \(p\mathrm{OH}\) = 14 - \(p\mathrm{H}\) = 14 - 12.39 = 1.61 We can find out the concentration of \(\mathrm{OH}^{-}\) by using \(p\mathrm{OH} = -log_{10}\ [\mathrm{OH}^{-}]\). The concentration of \(\mathrm{OH}^{-}\) = \(10^{-1.61}\) \(\mathrm{mol / L}\) This concentration of \(\mathrm{OH}^{-}\) can be attributed to the excess \(\mathrm{KOH}\). To find the volume of this excess \(\mathrm{KOH}\), we can use the relationship between volume and concentration: Volume of excess \(\mathrm{KOH}\) = (moles of excess \(\mathrm{KOH}\)) / (\(0.731 \mathrm{M}\)) First, we need to find the moles of excess \(\mathrm{KOH}\). Moles of excess \(\mathrm{KOH}\) = (volume of solution) \(\times\) (excess \(\mathrm{OH}^{-}\) concentration) Volume of solution = volume of \(\mathrm{HClO}_{4}\) + volume of \(\mathrm{KOH}\) = \(0.02500 \mathrm{L} + 0.04235 \mathrm{L} = 0.06735 \mathrm{L}\) Moles of excess \(\mathrm{KOH}\) = \(0.06735 \mathrm{L} \times 10^{-1.61}\ \mathrm{mol / L}\)= \(0.00417\ \mathrm{mol}\) Now, we can find the volume of this excess \(\mathrm{KOH}\): Volume of excess \(\mathrm{KOH}\) = \(0.00417\ \mathrm{mol} \ / \ 0.731\ \mathrm{M}\) = \(0.00571\ \mathrm{L}\) or \(5.71 \ \mathrm{mL}\) So, the student added \(5.71 \ \mathrm{mL}\) of \(\mathrm{KOH}\) in excess. The final answer: (a) Yes, the student went beyond the equivalence point. (b) The molarity of the strong acid (\(\mathrm{HClO}_{4}\)) is \(1.2368 \mathrm{M}\). (c) The student added \(5.71 \ \mathrm{mL}\) of \(\mathrm{KOH}\) in excess.

Key concepts

Understanding Acid-Base Titration

Acid-base titration is a method used to determine the concentration of an acid or a base in a solution. It involves the gradual addition of a solution of known concentration, called the titrant, to a solution of unknown concentration, called the analyte, until the reaction between the two is complete. This point of completion is signaled by a change in the property of the solution, such as its pH, which can be detected using an indicator or a pH meter.

In the context of the exercise, potassium hydroxide (KOH), a base, serves as the titrant, and perchloric acid (HClO₄), an acid, is the analyte. The student's ultimate goal is to find out how much KOH is required to neutralize the HClO₄. This process exemplifies an acid-base titration, where the reaction proceeds until all the acid has been neutralized by the base.

The Equivalence Point Explained

The equivalence point in a titration is the juncture at which the quantity of titrant added is stoichiometrically equivalent to the amount of analyte in the solution. In other words, it's the point at which there are equal moles of acid and base present, and the reaction between them is exactly complete.

Indicators and pH Monitoring

Typically, an indicator which changes color at a specific pH range is used to visually identify the equivalence point. Alternatively, as in our exercise, a pH meter can measure the precise pH level. The student in the exercise forgot to add an indicator, but by using a pH meter, they could still determine that the reaction had passed the equivalence point because the pH was above 7, which suggests a basic solution.

pH Calculation and Its Role in Titration

In the realm of chemistry, pH is a scale used to specify the acidity or basicity of an aqueous solution. It is a logarithmic scale based on the concentration of hydrogen ions (H⁺) in the solution. The pH scale typically ranges from 0 to 14, with pH 7 being neutral. For strong acids and bases, which completely dissociate in water, the pH or pOH can be calculated directly from their molar concentrations.

pH to Excess Base

In our exercise, the pH of the solution was 12.39 after the addition of KOH, indicating a basic solution. The student used this pH to calculate the pOH, which was then used to determine the concentration of excess hydroxide ions (OH^-). By finding the concentration of this excess base, it was possible to calculate how much KOH was added beyond the equivalence point. Note that the understanding of pH calculation is essential here, as it allows the determination of whether the titration has reached or exceeded the equivalence point without a visual indicator.