Problem 64 - Applied Statistics and Probability for Engineers

Question

In an acid-base titration, a base or acid is gradually added to the other until they have completely neutralized each other. Because acids and bases are usually colorless (as are the water and salt produced in the neutralization reaction), \(\mathrm{pH}\) is measured to monitor the reaction. Suppose that the equivalence point is reached after approximately \(100 \mathrm{~mL}\) of an \(\mathrm{NaOH}\) solution has been added (enough to react with all the acetic acid present) but that replicates are equally likely to indicate from 95 to \(104 \mathrm{~mL}\) to the nearest mL. Assume that volumes are measured to the nearest \(\mathrm{mL}\) and describe the sample space. (a) What is the probability that equivalence is indicated at \(100 \mathrm{~mL} ?\) (b) What is the probability that equivalence is indicated at less than \(100 \mathrm{~mL} ?\) (c) What is the probability that equivalence is indicated between 98 and \(102 \mathrm{~mL}\) (inclusive)?

Step-by-Step Solution

Answer

(a) \( \frac{1}{10} \); (b) \( \frac{1}{2} \); (c) \( \frac{1}{2} \).

Step 1: Identify the Sample Space

The range of possible outcomes (measured volumes, to the nearest mL) for equivalence in this titration is from 95 mL to 104 mL, inclusive. Therefore, the sample space can be listed as {95, 96, 97, 98, 99, 100, 101, 102, 103, 104}.

Step 2: Determine the Sample Space Elements Count

Count the total number of outcomes in the sample space: From 95 to 104 mL, this results in 10 possible outcomes (10 distinct mL readings).

Step 3: Calculate the Probability for 100 mL

The probability of any specific reading being indicated, including at 100 mL, is the ratio of favorable outcomes to the total number of outcomes. There is 1 favorable outcome, 100 mL, out of 10 possible outcomes: \( P(100) = \frac{1}{10} \).

Step 4: Calculate the Probability for Less than 100 mL

The readings less than 100 mL are 95, 96, 97, 98, 99 mL, giving us 5 favorable outcomes. Therefore, the probability is \( \frac{5}{10} = \frac{1}{2} \).

Step 5: Calculate the Probability for Between 98 and 102 mL (Inclusive)

Readings between 98 mL and 102 mL, inclusive, are 98, 99, 100, 101, 102, which amounts to 5 outcomes. Thus, the probability is also \( \frac{5}{10} = \frac{1}{2} \).

Key Concepts

Sample Space in Probability

When dealing with probability, a crucial first step is to define the sample space. The sample space is simply the set of all possible outcomes for an experiment or a trial. For example, when flipping a coin, the sample space is typically \( \{\text{Heads}, \text{Tails}\} \). In the context of our titration exercise, the sample space consists of all the possible measurements in milliliters where equivalence could be indicated. Since measures are rounded to the nearest mL, our sample space therefore includes every ml increment from 95 mL to 104 mL.

Understanding the sample space helps in calculating probabilities, as the probability of any given outcome is determined by counting the number of favorable outcomes and dividing by the total number of outcomes. In this titration experiment example, there are 10 potential outcomes in the sample space.

Titration Experiment

A titration experiment is a laboratory technique in chemistry where a solution of known concentration (the titrant) is gradually added to a solution of unknown concentration until the reaction reaches completion. This is typically to find the concentration of the unknown solution. Titrations are valuable in determining the neutralization point, which is most accurately achieved through careful measurement.

Throughout the process, careful monitoring is essential. Indicators such as pH change are often monitored to ascertain when neutralization or equivalence is reached. In our titration example, the titration process spans from a starting point until approximately 100 mL of the titrant has been added. However, slight variations are expected, which is why results range from 95 to 104 mL.

• Base is added slowly to an acidic solution.
• Molecular interactions occur until all acid is neutralized.
• Indicator, typically pH, signals the equivalence point.

Equivalence Point in Titration

In titration, reaching the equivalence point means that equal amounts of acid and base have reacted with each other. In other words, moles of acid equal the moles of base in the reaction. This is a crucial milestone because it indicates a complete reaction, often determined by a significant change in the solution's pH.

During a titration, the equivalence point can be identified several ways, including using a pH meter or an indicator dye that changes color when the pH reaches a certain level. In the case we're exploring, the equivalence point is shown as ranging between 95 and 104 mL, depending due to slight variances in each trial.

Knowing the equivalence point is key in calculating both the concentration of the unknown solution and for assessing the accuracy of the experiment. It's a cornerstone of calculations and helps in understanding the titrations' integrity.