Question

Consider an experiment in which two burets, Y and Z, are simultaneously draining into a beaker that initially contained 275.0 mL of 0.300 M HCl. Buret Y contains 0.150 M NaOH and buret Z contains 0.250 M KOH. The stoichiometric point in the titration is reached 60.65 minutes after Y and Z were started simultaneously. The total volume in the beaker at the stoichiometric point is 655 mL. Calculate the flow rates of burets Y and Z. Assume the flow rates remain constant during the experiment.

Short answer

The flow rates for burets Y and Z can be calculated by first finding the moles of HCl, NaOH, and KOH at the stoichiometric point. We find that initially, there are 82.5 mmol of HCl. At the stoichiometric point, we can write 82.5 mmol = Moles of NaOH + Moles of KOH. The total volume of NaOH and KOH at the stoichiometric point is 380 mL. By setting up and solving a system of two linear equations, we determine the volumes of NaOH (V_NaOH) and KOH (V_KOH). Finally, we calculate the flow rates of burets Y and Z by dividing the volumes of NaOH and KOH by the time (60.65 minutes). The flow rates are given as follows: Flow rate of buret Y \(= \frac{V_{NaOH}}{60.65}\) minutes Flow rate of buret Z \(= \frac{V_{KOH}}{60.65}\) minutes

Step-by-step solution

Calculate the moles of HCl present at the beginning

To calculate the moles of HCl in the beaker initially, we can use the formula: Moles of HCl = Molarity × Volume Given, the initial volume of HCl is 275.0 mL and its molarity is 0.300 M. Moles of HCl = 0.300 M × 275.0 mL = 82.5 mmol

Find the moles of NaOH and KOH at the stoichiometric point

At the stoichiometric point, the moles of HCl are completely neutralized by the moles of NaOH and KOH. Since the reaction between HCl, NaOH, and KOH is 1:1, we can write: Moles of HCl = Moles of NaOH + Moles of KOH 82.5 mmol = Moles of NaOH + Moles of KOH

Calculate the total volume of NaOH and KOH at the stoichiometric point

Given, the total volume in the beaker at the stoichiometric point is 655 mL. The initial volume of HCl was 275.0 mL, so the combined volume of NaOH and KOH would be: Total volume of NaOH and KOH = 655 mL - 275 mL = 380 mL

Determine the volumes of NaOH and KOH required to reach the stoichiometric point

We can use the given concentrations and the moles of NaOH and KOH from Step 2 to find the volumes of NaOH and KOH at the stoichiometric point. Let the volume of NaOH be V_NaOH, and the volume of KOH be V_KOH. (0.150 M × V_NaOH) + (0.250 M × V_KOH) = 82.5 mmol We know the sum of these volumes should be 380 mL (from step 3): V_NaOH + V_KOH = 380 mL We now have a system of two linear equations with two variables: 0.150 × V_NaOH + 0.250 × V_KOH = 82.5 V_NaOH + V_KOH = 380 Use a suitable method to solve this system to find V_NaOH and V_KOH.

Calculate the flow rates of burets Y and Z

Now that we have the volumes of NaOH and KOH at the stoichiometric point, we can find the flow rates for each buret. Since the flow rates are assumed to be constant during the experiment, we can divide the volume by the time (in minutes) to obtain the flow rate. Flow rate of buret Y (containing NaOH): (V_NaOH) / 60.65 minutes Flow rate of buret Z (containing KOH): (V_KOH) / 60.65 minutes Calculate the flow rates for burets Y and Z to get the final answer.

Key concepts

Stoichiometric Point

The stoichiometric point is an essential concept in titration. It is the point at which the amount of titrant added exactly neutralizes the analyte in the solution. In other words, the reactants have completely reacted according to the reaction's stoichiometry.
In the case of acid-base titrations, like the one in the exercise, reaching the stoichiometric point means that the moles of acid are equal to the moles of base added. This exact balance is crucial for calculating other parameters, like the flow rates of burets, as seen in the original solution.
Recognizing the stoichiometric point is vital to ensure precise, error-free calculations and outcomes in any titration process.

Neutralization Reaction

A neutralization reaction is a chemical reaction where an acid and a base react to form water and a salt. Typically, this is represented by the equation:
\[ ext{Acid + Base} ightarrow ext{Salt + Water} \]
In the provided experiment, hydrochloric acid (HCl) is neutralized by two bases: sodium hydroxide (NaOH) and potassium hydroxide (KOH). The interaction between the acid (HCl) and bases results in water and salts, NaCl and KCl, respectively.
Neutralization reactions are usually exothermic, releasing heat. As these reactions reach completion at the stoichiometric point, they play a pivotal role in determining the volume of titrant needed. Understand that in this experiment, both bases contribute cumulatively to neutralizing the acid, making their careful measurement critical for accuracy.

Molarity

Molarity is a concentration unit expressed as moles of solute per liter of solution. It allows chemists to quantify the concentration of a substance in a given volume of liquid. The formula for molarity (\( M \)) is:
\[ M = \frac{n}{V} \]
where \( n \) is the number of moles of solute and \( V \) is the volume of the solution in liters.
In the exercise, the initial molarity of HCl is used alongside its volume to calculate the moles of HCl present initially. This is a typical starting point in titration calculations, as it sets the stage for understanding how much base is needed to achieve neutralization.
Proper grasp of molarity and its application is necessary to predict how substances will interact during a reaction and to prepare solutions with precise concentrations.

Flow Rate Calculation

Flow rate calculation is crucial when dealing with experiments that involve liquids being added or removed at a constant rate, like in this titration experiment. The flow rate tells you how fast a liquid is being transferred per unit time.
For the experiment, we calculate the flow rates of the two burets by dividing the volume of solution dispensed by the time taken. This is expressed by the formula:
\[ ext{Flow Rate} = \frac{ ext{Volume dispensed}}{ ext{Time}} \]
The flow rates are vital for ensuring precision and accuracy in titrations, especially since any variation could lead to incorrect stoichiometric conditions. Steady, predictable flow rates allow for more controlled experiments and reliable results.
Understanding this concept can help in troubleshooting experimental anomalies and ensuring consistency in replicated tests.