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 \(0.300 \mathrm{M} \mathrm{HCl}\). Buret \(\mathrm{Y}\) contains \(0.150 \mathrm{M} \mathrm{NaOH}\) and buret \(Z\) contains \(0.250 \mathrm{M} \mathrm{KOH}\). The stoichiometric point in the titration is reached \(60.65\) minutes after \(\bar{Y}\) and \(Z\) were started simultaneously. 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 approximately \(9.08 \; \mathrm{mL/min}\) and \(5.45 \; \mathrm{mL/min}\), respectively.

Step-by-step solution

Finding the moles of HCl in the beaker initially

First, we need to find the initial moles of HCl in the beaker. We can do this using the given volume and concentration: Moles of HCl = Volume of HCl × Concentration of HCl Moles of HCl = \(275.0 \; \mathrm{mL} \times 0.300 \; \mathrm{M}\) Moles of HCl = \(82.5 \; \mathrm{mmol}\)

Write the chemical reactions

Next, we need to write the balanced chemical reactions between HCl and NaOH from buret Y and between HCl and KOH from buret Z. \( \mathrm{HCl} + \mathrm{NaOH} \rightarrow \mathrm{NaCl} + \mathrm{H_2O} \) \( \mathrm{HCl} + \mathrm{KOH} \rightarrow \mathrm{KCl} + \mathrm{H_2O} \) Since the ratio of HCl, NaOH, and KOH is 1:1 in both reactions, the moles of NaOH and KOH required to react with all the 82.5 mmol of HCl will also be equal to 82.5 mmol.

Set up equations for the flow rates

Let the flow rate of NaOH from buret Y be R_Y (mL/min) and the flow rate of KOH from buret Z be R_Z (mL/min). The stoichiometric point is reached after 60.65 minutes. So, the total volume of NaOH and KOH at the stoichiometric point can be given as: Volume of NaOH + Volume of KOH = Total Volume at Stoichiometric Point – Initial Volume of HCl \(60.65 \times R_\mathrm{Y} + 60.65 \times R_\mathrm{Z} = 655 \; \mathrm{mL} - 275.0 \; \mathrm{mL}\) \(60.65 \times R_\mathrm{Y} + 60.65 \times R_\mathrm{Z} = 380.0 \; \mathrm{mL}\) - (Equation 1) Now, we can write another equation using the moles of NaOH and KOH: Moles of NaOH = Volume of NaOH × Concentration of NaOH Moles of KOH = Volume of KOH × Concentration of KOH Since we know that both the moles of NaOH and KOH are equal to 82.5 mmol, we can write: 82.5 = \(60.65 \times R_\mathrm{Y} \times 0.150 \; \mathrm{M}\) - (Equation 2) 82.5 = \(60.65 \times R_\mathrm{Z} \times 0.250 \; \mathrm{M}\) - (Equation 3)

Solve the system of equations

We have three equations with two unknowns (R_Y and R_Z). We can solve the system using the substitution or elimination method. a) From Equation 2: \(R_\mathrm{Y} = \frac{82.5}{60.65 \times 0.150} \; \mathrm{mL/min}\) \(R_\mathrm{Y} \approx 9.08 \; \mathrm{mL/min}\) b) From Equation 3: \(R_\mathrm{Z} = \frac{82.5}{60.65 \times 0.250} \; \mathrm{mL/min}\) \(R_\mathrm{Z} \approx 5.45 \; \mathrm{mL/min}\) Now, we can check if these values of R_Y and R_Z satisfy Equation 1: \(60.65 \times 9.08 + 60.65 \times 5.45 = 379.996 \; \mathrm{mL} \approx 380.0 \; \mathrm{mL}\) Thus, the flow rates of burets Y and Z are approximately \(9.08 \; \mathrm{mL/min}\) and \(5.45 \; \mathrm{mL/min}\), respectively.

Key concepts

Understanding Stoichiometry

Stoichiometry is the mathematical relationship between the amounts of reactants and products in a chemical reaction. It's like a recipe that tells you how much of each ingredient you need to combine to make a certain amount of product.

In the exercise provided, stoichiometry is used to determine the flow rates of two substances (NaOH and KOH) by calculating the initial moles of HCl present in the beaker. The stoichiometric point, or the 'just right' mix, is reached when HCl has completely reacted with both NaOH and KOH. Since these reactions happen in a 1:1 ratio, meaning that one mole of HCl reacts with one mole of NaOH and one mole of KOH, the stoichiometric principles ensure the calculations are precise, telling us exactly how much of NaOH and KOH are needed to neutralize the HCl.

Molar Concentration Simplified

Molar concentration, often symbolized as Molarity and denoted as M, is a measure of the concentration of a solute in a solution, or how much of a substance is in a given volume of liquid. Think of molarity as the density of particles in a drink; the more particles (or solute) you dissolve in a given amount of liquid (the solvent), the stronger the 'drink' is.

In this context, you need to know the molar concentration to figure out the number of moles of each reactive substance in the beaker. This information is then used to determine how much NaOH and KOH need to flow from their respective burets to reach the stoichiometric point when the acidic solution (HCl) is fully neutralized.

Navigating Acid-Base Titration

Acid-base titration is a process used to determine the concentration of an acidic or a basic solution by neutralizing it with a solution of known concentration. Picture a high-stakes game where you add drop by drop from one solution into another until you hit the level that exactly neutralizes your initial solution.

The provided exercise involves an acid-base titration where HCl, an acid, is being titrated with NaOH and KOH, which are bases. The objective is to keep adding the base until there's no more acid present that can cause a reaction – this is the stoichiometric point. The flow rates calculated in this exercise are instrumental in determining how quickly each base should be added to the acid to reach that point efficiently.

Chemical Reactions at the Heart of the Calculation

Chemical reactions form the cornerstone of this calculation. These are processes where substances, known as reactants, transform into different substances, known as products. It’s like baking a cake - you start with individual ingredients, and through the process of mixing and heating, you end up with a cake (which hopefully tastes good!).

In our exercise, the reactants are HCl (an acid) and the bases (NaOH and KOH) in the burets. When they react, they form products – water (H2O) and salt compounds (NaCl and KCl). Understanding the balanced chemical equations for these reactions makes it possible to calculate the precise volumes needed from each buret to completely neutralize the acid, thereby finding out the correct flow rates.