Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

How many milliliters of concentrated hydrochloric acid solution (36.0% HCl by mass, density =1.18g/mL) are required to produce 10.0 L of a solution that has a pH of 2.05?

Short Answer

Expert verified
Approximately 7.69 mL of concentrated hydrochloric acid solution (36.0% HCl by mass, density = 1.18 g/mL) is required to produce 10.0 L of a solution with a pH of 2.05.

Step by step solution

01

Determine the concentration of HCl in the final solution

To find the concentration of HCl in the final solution, we use the formula for pH: pH = -log[H+]. In our case, we are given a pH value of 2.05, and we have to find the concentration of hydrogen ions [H+], which in the case of HCl solution equals [HCl]. So: 2.05=log([HCl])
02

Calculate the concentration of HCl

Now we solve for the concentration [HCl]. Take the inverse of the log using 10^(−pH): [HCl]=102.05\]Calculatetheconcentration[HCl]:\([HCl]8.91×103M
03

Find the moles of HCl required for the final solution

We now have the concentration of HCl in the final solution, 8.91×10^(-3) M. The volume of the final solution is 10 L. We can find the moles of HCl required for this solution using the formula: Moles of HCl = Concentration × Volume Moles of HCl = 8.91×103M×10L Calculate the moles of HCl: Moles of HCl ≈ 0.0891 moles
04

Calculate the mass of HCl required

The mass of HCl required can be found by multiplying the moles of HCl by its molar mass (36.5 g/mol), as given in the problem: Mass of HCl = Moles of HCl × Molar mass Mass of HCl = 0.0891moles×36.5g/mol Calculate the mass of HCl: Mass of HCl ≈ 3.25 g
05

Calculate the volume of concentrated HCl required

The concentrated HCl solution is given as 36% by mass with a density of 1.18 g/mL. To find the volume of concentrated HCl required, we can use the mass of HCl required, the percentage composition of HCl in the concentrated solution, and the density of the concentrated solution: Volume of concentrated HCl = (Mass of HCl required) / (Percentage composition × Density) Volume of concentrated HCl = (3.25g)/(0.36×1.18g/mL) Calculate the volume of concentrated HCl: Volume of concentrated HCl ≈ 7.69 mL So, approximately 7.69 mL of concentrated hydrochloric acid solution is required to produce 10.0 L of a solution with a pH of 2.05.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

pH Calculation
In chemistry, the pH scale is crucial for determining the acidity or basicity of a solution. pH is defined as the negative logarithm of the hydrogen ion concentration pH=log[H+]. This scale ranges from 0 to 14, where pH 7 is neutral, below 7 is acidic, and above 7 is basic. For hydrochloric acid (HCl), a strong acid, the pH directly correlates to the concentration of HCl itself because it dissociates completely in water.

To understand this concept better, consider the task at hand, where the given pH is 2.05. Using the relationship pH=log[H+], you can find the concentration of HCl as 102.05 which approximately equals 8.91×103M. This calculation shows how you can determine the concentration from the known pH, a vital skill in various chemical applications.

Molarity
Molarity, often denoted as M, is the concentration of a solute in a solution, expressed as moles of solute per liter of solution. In other words, it's a measure of how many molecules of a substance are present in a given volume of liquid.

To find the molarity of HCl in this specific problem, we use the equation [HCl]=8.91×103M obtained from the pH calculation. This value signifies how concentrated the acid is in the solution.


More clearly, if you have a 10-liter solution and the concentration is [HCl]=8.91×103M, it means you have 8.91 millimoles of HCl in every liter of your solution. In practical terms, this concentration gives insight into how much of the solute you'll be working with in laboratory settings or chemical reactions.
Density
Density is a physical property that measures the mass per unit volume of a substance, typically expressed in g/mL or kg/L. It can help identify substances and predict how they will interact in various scenarios.

In the exercise, the given density of concentrated hydrochloric acid is 1.18 g/mL. This figure indicates that every milliliter of the hydrochloric acid solution weighs 1.18 grams.


Density is crucial when converting between mass and volume, especially when forming solutions. By using density, you can calculate how much volume of a concentrated acid you need to achieve a desired mass of solute. As shown in this problem, density helps determine the volume of concentrated hydrochloric acid needed to achieve the desired concentration in the final solution.
Concentration
Concentration expresses how much solute is present in a given quantity of solvent or solution. It can be expressed in various units, but molarity, mass percent, and molality are the most common.

In this problem, concentration comes into play primarily in two forms: as the molarity of the final solution 8.91×103M and as the mass percentage of the concentrated HCl solution 36%.


The mass percentage indicates that in 100 grams of the concentrated solution, 36 grams consist of HCl. Understanding concentration is fundamental for safely and efficiently preparing solutions. It ensures the accuracy of chemical reactions and lab procedures, where incorrect concentrations can lead to poor results or even hazardous situations. Thus, being able to translate between different concentrations is an essential skill in chemistry.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.

Sign-up for free