February 14th - Solution Stoichiometry

Today, we started our new unit: solution stoichiometry! In this unit, we’ll learn how to apply our knowledge of stoichiometry to solution chemistry. We’ll do a few examples, but first, we need to review some key points.

Solutions are homogeneous mixtures composed of a solute and a solvent. If you aren’t familiar, solutes are the chemicals in lesser amounts in a reaction. In a solution, it is whatever dissolves. Solvents are the chemicals present in a great amount, or whatever does the dissolving.

Here is a picture of acetic acid, a common solvent.

As we already know, chemicals dissolved in water are aqueous. They have an _(aq) symbol (ex. NaCl_(aq)). If you want to show a reaction for something dissolved in water, you don’t include water on either side of the ‘reacts to form’ (à) symbol . Instead, it is shown by putting HOH above the à.

This equation shows Sodium chloride dissolving in water.

Concentration of a chemical in a reaction can be expressed in many ways. Grams per litre (g/L), percent by volume, percent by mass, and moles per litre (mol/L) are all acceptable. The most common is mol/L, which is called molarity. It is also important to note that mol/L is often replaced by the symbol M.

Molarity is an expression of the number of moles of the solute divided by the volume of the solvent. When you see the chemical symbol of a compound or element in square brackets, such as [HCl], it is referring to the concentration of said compound/element.

With all the notes out of the way, we can do a few examples!

Ex.) What is the concentration of NaCl_(aq) made from 1.2 mol of the chemical dissolved in 0.75 L of water?

Using quantitative analysis, we must convert 1.2 mol in .75 L to M, or mol/L. We do so like this:

1.2 mol   x   (1/.75 L)   =   1.6 mol/L, or 1.6 M

If we know the number of moles and the volume, we could simply divide them. However, it is better to set up the equation like I just did. When we get into trickier questions later in the unit, it will make things easier.

Diluted solutions would have low molarity, while concentrated ones would have high molarity.

Ex.) Determine the mass of AgNO₃ required to create 0.100 M in 250 mL of water.

We use quantitative analysis yet again. Keep in mind that 250 mL is equal to 0.250 L. Set up the equation like so:

0.250 L   x   (0.100 mol/L)   =   0.0250 mol of AgNO₃

We then simply convert moles to mass using the molar mass of AgNO₃.

0.0250 mol   x   (169.9g/mol)   =   4.25 g of AgNO₃

Ex.) How many litres of water are required to make a solution when 0.250 mol is dissolved to create a 1.75 M solution of Na₂S?

This time, we start with moles first and convert to how many litres of water we need. Simple!

0.250 mol   x   (L/1.75 mol)   =   0.143 L of HOH

The question seems daunting, but the procedure is simple. The hardest part of solution stoichiometry is determining the order in which to convert.

As always, here is a superb video on solution stoichiometry. Enjoy!

Posted by Michael.