Question

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

Short answer

The flow rates of burets Y and Z are 3.96 mL/min and 2.31 mL/min, respectively.

Step-by-step solution

Calculate the moles of HCl initially present

We can calculate the moles of HCl initially present in the beaker using the formula: moles = volume × concentration The initial volume of HCl is 275.0 mL, and its concentration is 0.300 M. Moles of HCl = \( 275.0 \times 0.300 \) mL.M = 82.5 mmol

Determine the moles of OH⁻ ions needed for the stoichiometric point

At the stoichiometric point, we need an equal number of moles of OH⁻ ions to react with HCl. Therefore, the total moles of OH⁻ ions needed are: Total moles of OH⁻ = 82.5 mmol

Find the total volume of OH⁻ ions needed

We know that the stoichiometric point is reached when the total volume in the beaker is 655 ml. Thus, the combined volume of NaOH and KOH solutions added is: Volume of NaOH + Volume of KOH = 655 - 275 = 380 mL

Set up and solve the system of equations

Let V_Y be the volume of NaOH from buret Y and V_Z be the volume of KOH from buret Z. Using the concept of moles and concentration, we have the following system of equations: Equation 1: V_Y + V_Z = 380 mL (total volume of NaOH and KOH) Equation 2: \(0.150 \times V_Y\) + \( 0.250 \times V_Z\) = 82.5 mmol (total moles of OH⁻) Now, we can solve the equations by substituting V_Z from equation 1 into equation 2: \( 0.150 \times V_Y\) + \( 0.250 \times (380 - V_Y) = 82.5 \) Solving for V_Y, we get: V_Y = 240 mL Now substituting the value of V_Y into equation 1, V_Z = 380 - 240 = 140 mL

Calculate the flow rates

Since the stoichiometric point is reached after 60.65 minutes, we can find the flow rates of Y and Z as follows: Flow rate of Y = \( \frac{V_Y}{time} = \frac{240}{60.65} \) = 3.96 mL/min Flow rate of Z = \( \frac{V_Z}{time} = \frac{140}{60.65} \) = 2.31 mL/min So, the flow rates of burets Y and Z are 3.96 mL/min and 2.31 mL/min, respectively.

Key concepts

Titration

Titration is a common laboratory technique used to determine the concentration of an unknown solution. It involves slowly adding a solution of known concentration, called the titrant, to the unknown solution, until the reaction between the two solutions is complete. This point of completion is known as the equivalence or stoichiometric point. In a typical acid-base titration, the stoichiometric point is reached when the number of moles of hydrogen ions (H⁺) in the acid equals the number of moles of hydroxide ions (OH^-) in the base. Indicators or pH meters are commonly employed to signify the reaction's endpoint visually.

Application in the Exercise

In the given exercise, a titration is performed using two burets containing different base solutions, NaOH and KOH, to neutralize an HCl solution. As each drop of base is added, it reacts with the acid until the moles of HCl are equal to the combined moles of OH^- from NaOH and KOH, indicating the stoichiometric point.

Stoichiometry

Stoichiometry is the branch of chemistry that quantitatively relates the amounts of reactants and products in a chemical reaction. It is grounded in the law of conservation of mass and relies on the principle that during a chemical reaction, atoms are neither created nor destroyed, only rearranged.
Stoichiometry calculations involve using the coefficients from the balanced chemical equations to convert between moles, mass, and particles of reactants and products. Understanding stoichiometry allows chemists to predict the quantities needed or produced in a given reaction, making it a crucial aspect of chemistry.

Application in the Exercise

In our exercise, stoichiometry is employed to calculate the moles of hydroxide ions required to neutralize the given moles of hydrochloric acid. Using this stoichiometric relationship, the total volume of NaOH and KOH that must be added to reach the stoichiometric point can be determined.

Molarity

Molarity, denoted as M, is a measure of concentration for solutions, defined as the number of moles of solute per liter of solution. It is one of the most commonly used units of concentration in chemistry. When preparing a solution or performing a reaction, knowing the molarity is important as it describes how much substance is present in a given volume of liquid.
To calculate molarity, you can use the formula:
\[ M = \frac{\text{moles of solute}}{\text{liters of solution}} \]
Molarity is a central concept in titrations, as it allows chemists to determine how solutions of a known concentration will react with a given volume of another solution.

Application in the Exercise

In our scenario, the molarity of each solution (NaOH, KOH, and HCl) is given, and it is utilized, along with the volumes of the solutions from the burets, to calculate the moles of reactants involved at the stoichiometric point.

Flow Rate Calculation

Flow rate is the volume of liquid that is dispensed over a given period of time. It is often measured in units like milliliters per minute (mL/min) in chemistry experiments involving liquids. Accurately determining the flow rate is essential in processes such as titration, where the rate at which the titrant is added can affect the accuracy of the results.
In practical terms, to calculate the flow rate, you can use the formula:
\[ \text{Flow rate} = \frac{\text{Volume dispensed}}{\text{Time taken}} \]
This calculation is important for ensuring precise control over chemical reactions and processes.

Application in the Exercise

The flow rates for burets Y and Z in our exercise are computed using the volumes of NaOH and KOH that were dispensed to reach the stoichiometric point and the time elapsed during the experiment. These calculated flow rates are crucial for ensuring that the bases were added at a constant and controlled rate throughout the titration.