Acids and Bases, and How to Calculate Their Molality?

Acids and Bases

Acids have long been recognised as a distinct class of compounds with the following aqueous solutions properties:

  • A distinct sour flavour.
  • The colour of litmus is changed from blue to red.
  • When it comes into contact with certain metals, it produces gaseous H2.
  • When combined with bases, it produces salt and water.
  • Acidic solutions have a pH of less than 7, with lower pH values indicating greater acidity.
  • Acids include acetic acid (found in vinegar), sulfuric acid (found in car batteries), and tartaric acid (used in baking).

On the other hand, bases can be recognised as a distinct class of compounds with the following properties.

  • They have a bitter taste.
  • They are soapy to the touch.
  • They change the colour of the litmus from red to blue.
  • Basic solutions have a pH of more than 7, with lower pH values indicating a weaker base.
  • Most alkali metal and some alkaline earth metal hydroxides are strong bases in the solution. These are some examples: Sodium hydroxide (NaOH), potassium hydroxide (KOH), lithium hydroxide (LiOH), and rubidium hydroxide (RbOH).

Acid and base are important concepts that we encounter or study in chemistry. Furthermore, acid and base are two major classifications of substances that include both laboratory chemicals and substances found in everyday life.

To define acids and bases, three different theories have been proposed: The Arrhenius theory, the Bronsted-Lowry theory, and the Lewis theory of acids and bases.

  • According to the Arrhenius theory of acids and bases, “an acid generates H+ ions in a solution, whereas a base generates an OH ion in its solution.”
  • According to the Bronsted-Lowry theory, “an acid is a proton donor, and a base is a proton acceptor.”
  • Finally, the Lewis definition of acids and bases states that “acids are electron-pair acceptors, while bases are electron-pair donors.”

Molality of Acids and Base

A solution’s molality is calculated by dividing the moles of solute by the kilograms of solvent. A lowercase “m” represents molality. We commonly submit concentrations in molality because, unlike molarity, molality is not temperature-dependent. This autonomy makes it easier for scientists worldwide to replicate the work.

Molality is primarily used to express the concentrations of solutions in relation to vapour pressure and temperature changes. Molality is also used to determine the boiling or melting point, as well as when working with colligative properties. In terms of calculation, molality can be easily determined by knowing the masses of solute and solvent in a solution. Furthermore, the concentration or molality of a homogeneous solution is always constant.

Example-

  1. Find the molality of 2.5 M H2SO4. This solution has a density of 1.54 g/mL.

Step 1: Let us make an assumption. Let’s say you have 1 L of solution. This is a critical step, and the amount of solution is not specified, but you must have a specific quantity to perform the calculations, and one litre is the best guess for this problem.

Step 2: Calculate the total mass of the solution. Multiply 1 L by the density (1.54 g/mL) multiplied by 1000 mL/L. This yields 1540 g of solution.

Step 3: Determine the solute’s gram weight. 2.5M is the equivalent of 2.5 moles of sulphuric acid per litre of solution. 2.5 moles equals grams. Sulphuric acid has a molar mass of 98.09 g/mol. 2.5 moles multiplied by 98.09 g/mol equals 245.225 grams of sulphuric acid.

Step 4: Determine the weight of the solvent. 1540 g solution – 245.225 g solute = 1294.775 g or 1.294775 kg solvent

Step 5: Determine the molality. 2.5 moles solute / 1.294775 kg solvent = 1.9 molal H2SO4.

  1. What is the molality of a 1M NaOH solution?

Step 1: Let us make an assumption. Assume you have one litre of solution. Then there’s 1 mol (40 g) of NaOH.

The solution has a density of 1.04 g/mL.

Step 2: Calculate the total mass of the solution. Mass of solution = 1040 g solution.

Step 3: Determine the mass of water = (1040 – 40) g = 1000 g = 1.0 kg

Step 4: Determine the molality.

m = moles of solute / kilogramme of solvent = 1 mol/1.0 kg= 1 mol/kg

1 mol/kg is the molal concentration.

Leave a Reply

Your email address will not be published.

© 2022 Nor Start - WordPress Theme by WPEnjoy