SOLUTIONS: Are homogeneous mixtures.
SOLVENT: The chemical in bigger proportion.
SOLUTE: The chemical in lower proportion.
We will concentrate our attention in the AQUEOUS SOLUTIONS since these are the ones that we use in the chemistry lab most of the time. Aqueous solutions are the ones in which the solvent is water. We can also find ALCOHOLIC SOLUTIONS, where the solvent is alcohol.
ELECTROLYTES: If the solute liberates ions in solution, (Ionic compounds that separate in their ions or acids that release hydrogen ions in solution) those ions will allow the solution to conduct electricity. These substances are called electrolytes
To prepare STANDARD SOLUTION in chemistry, we use a VOLUMETRIC FLASK
To prepare a standard solution of a specific concentration we need to take the following steps:
- calculate the correct mass of the solute needed.
- weigh the solute with an analytic balance (digital)
- dissolve the solute in some DISTILLED WATER using a beaker.
- using a funnel, transfer the liquid into the volumetric flask.
- Wash the beaker and funnel into the volumetric flask to be sure that all the solute was transferred into the flask
- using a stopper, cover the volumetric flask and move it gently several times to ensure that the solute is distributed evenly.
- adjust the volume of the solution
- using a stopper, cover the volumetric flask and move it gently again several times to ensure that the solute is distributed evenly.
There are several reasons why we use solutions instead of pure chemicals during experiments or labs.
- Solid chemicals cannot move freely and the collisions needed for the reaction to occur will not happen.
- The chemicals contained in solutions will collide faster and better so the chances for the reaction to occur are bigger.
- The water used in the solutions absorbs some of the heat produced by the reaction and it is safer
Concentration of a solution:
Concentration: is a measurement of how much solute is present in the solution.
- The more solute a solution has per unit of volume, the more CONCENTRATED it is.
- The less solute it has per unit of volume, the more DILUTED it is.
This is easy to view when we work with colorful compounds, but if we work with white compounds, they will form colorless solutions that cannot be seen easily.
A microscopic view:
When the solute dissolves, the particles will locate in between the solvent molecules (see graph below).
Solvent molecules move all the time, but the amount of free spaces determine how much solute will be able to dissolve in the liquid.
Based on the amount of solute and solvent a solution can be:
- UNSATURATED SOLUTION: There are free spaces available to receive more solute.
- SATURATED SOLUTION: This kind of solution will be formed when the solute occupies all free spaces in between the solvent molecules.
- SUPER SATURATED SOLUTION: is the one that has more solute than the maximum amount specified for that temperature. Sometimes this solution will stay like this if it is not disturbed.
- Example, If we heat up the liquid or we give them energy by swirling or shaking, it will create a temporary state with extra spaces. as soon as the energy get back to normal, those extra spaces will disappear and we will be able to see some solute at the bottom of the container.
CONCENTRATION OF A SOLUTION
MEASURING THE AMOUNT OF SOLUTE: “Concentration”
The concentration of a solution can be expressed in different units. We will use only one: MOLARITY
Molarity: It’s a measure of the concentration that express the NUMBER OF MOLES OF SOLUTE in each LITER OF SOLUTION
# of moles = volume (dm3) x concentration (M)
Important equivalences that you need to remember:
1000 cm3 = 1000 mL = 1 L = 1 dm3
# of moles = mass (g) / Mr (g/mol)
1) What is the molarity of a solution that contains 0.200 moles of NaCl in 0.500 dm3 of solution?
# moles of solute= 0.200 moles
volume of solution = 0.500 dm3
M = # moles of solute = 0.200 mol = 0.4 mol. dm-3
volume of solution (dm3) 0.500 dm3
2) What is the molarity of a solution that contains 0.500 moles of CuCl2 in 250 cm3 of solution?
# moles of solute= 0.500
volume of solution = 250 cm3= 0.250 dm3 (we cannot use cm3)
M = # moles of solute = 0.500 mol = 2.00 mol. dm-3 volume of solution (dm3) 0.250 dm32
3) What is the molarity of a solution in which 0.45 grams of sodium nitrate (NaNO3)are dissolved in 265 mL of solution.
- Calculate how many moles are there in the 0.45 g of sodium nitrate in the 265 mL of solution.
X mol NaNO3 = 1 mol NaNO3 X= 5.29 x 10 -3 moles
0.45 g 85 g
- Calculate the # of moles in 1 L
5.29 x 10 -3 moles = X moles X= 0.02 moles/L
0.265 L 1
4) Explain how to make one liter of a 1.25 molar ammonium Chloride solution. (NH4Cl)
- Calculate the molar mass of ammonium hydroxide = 53.5 g/mol
- Dissolve 66.875 grams (1.25 moles) of ammonium chloride in some water using a beaker
- Wash the beaker content into a 1 L (= 1 dm3) volumetric flask
- Adjust the volume up to the mark level.
- Mix thoroughl
5) Calculate the amount of moles of KI necessary to to prepare the 125 mL of solution 0.050 M
If the solution is 0.050M it means that there are 0.050 moles in a liter of solution, in 0.125L there will be
0.050 M = X ==> X= 6.25 x 10-3moles
1 L 0.125 L
6) Calculate the amount of grams of NaCl necessary to to prepare the 125 mL of solution 0.100 M
- calculate the amount of moles
- pass those moles to grams
1) If the solution is 0.050M it means that there are 0.050 moles in a liter of solution, in 0.125L there will be
# moles = M (mol/L) x Vol (L) ==> X= 0.0125 moles of NaCl
2) Calculate the mass of solute (NaCl) contained in the moles calculated in #1
0.0125 moles NaCl x 58.5 g NaCl ==> X= 0.731 g NaCl
1 mol NaCl
Solutions are prepared by diluting a more concentrated solution. For example, if you needed a one molar solution you could start with a six molar solution and dilute it.
Consequently, you also need to be familiar with the calculations that are associated with dilutions.There is an element of simplicity in calculations of these types and the element of simplicity is that the number of moles of solute stays the same, as shown here. The number of moles of solute in the concentrated solution (indicated by the subscripted molescon) is equal to the number of moles in the dilute solution. You have simply increased the amount of solvent in the solution.
# molesconcentrated solution = #molesdiluted solution
Of course you know that the number of moles of solute in the concentrated solution is equal to the molarity of the concentrated solution times the volume of the concentrated solution. Also, the number of moles of solute in the dilute solution is equal to the molarity of the dilute solution times the volume of the dilute solution. Since we are really interested in the molarities and volumes we can substitute and use the equation shown in the second line (or third line, depending on your preference for how to show multiplication). Let’s use this equation in a few examples.