Fascinating topics on Chemistry. Related posts;

Fascinating topics on Chemistry.


Strong Acid Definition and Examples;

 

A strong acid is one that is completely dissociated or ionized in an aqueous solution. It is a chemical species with a high capacity to lose a proton, H+. In water, a strong acid loses one proton, which is captured by water to form the hydronium ion:
 
Solvent: In most applications, strong acids are discussed in relation to water as a solvent. However, acidity and basicity have meaning in nonaqueous solvent. For example, in liquid ammonia, acetic acid ionizes completely and may be considered a strong acid, even though it is a weak acid in water.
Calculating the Concentration of a Chemical Solution;
 
Concentration is an expression of how much solute is dissolved in a solvent in a chemical solution. There are multiple units of concentration. Which unit you use depends on how you intend to use the chemical solution. The most common units are molarity, molality, normality, mass percent, volume percent, and mole fraction. Here are step-by-step directions for calculating concentration, with examples.
How to Calculate Molarity of a Chemical Solution:
Woman holding glass containing green liquid
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Molarity is one of the most common units of concentration. It is used when the temperature of an experiment won’t change. It’s one of the easiest units to calculate.
Calculate Molarity: moles solute per liter of solution (not volume of solvent added since the solute takes up some space)
symbol: M
M = moles / liter
Example: What is the molarity of a solution of 6 grams of NaCl (~1 teaspoon of table salt) dissolved in 500 milliliters of water?
First, convert grams of NaCl to moles of NaCl.
From the periodic table:
Na = 23.0 g/mol
Cl = 35.5 g/mol
NaCl = 23.0 g/mol + 35.5 g/mol = 58.5 g/mol
Total number of moles = (1 mole / 58.5 g) * 6 g = 0.62 moles
Now determine moles per liter of solution:
M = 0.62 moles NaCl / 0.50 liter solution = 1.2 M solution (1.2 molar solution)
Note that I assumed dissolving the 6 grams of salt did not appreciably affect the volume of the solution. When you prepare a molar solution, avoid this problem by adding solvent to your solute to reach a specific volume.
 
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