Transcript for:
Understanding Dilutions in Solutions

in Section 4.5 we will be discussing about dilutions of solutions okay let's say in the beginning you know given a beaker of concentrated solution concentrated solution sometimes is called a stock solution and then your task to prepare a diluted solution from the concentrate that one how can you do so we can prepare the diluted solution from the concentrate that one by simply adding solvent onto it all right so this process of preparing the diluted solution from the concentrated one is called a dilutions process in the dilution process it is very important to note that the amount of solute remains the same during the whole process and this leads us to the second page okay let's say solution one is the concentrated solution and solution two is the dilute at one right so as I mentioned earlier on in the whole dilution process the number of moles of solute remains the same alright so this is what we can do the number of moles of solute in the concentrated solution equals the number of solute in the dilute at one how do we calculate the number of moles of the solute we have to use this formula which you already know number of moles equals concentration times volume so the number of moles of solute in the concentrated solution is simply c1 times v1 and the number of moles of solute in the dilute that one is simply c2 times v2 all right and then this formula is very useful and it can help you to figure out the volume or the concentration of the concentrated solution that you need to prepare a dilute solution of certain concentration and volume alright let's look at a couple of examples the left officer wants to prepare 1.50 liters of 0.2 m HCL from a stop solution which sometimes we call concentrated solution of the commercial se which is twelve point zero and what volume of the concentrated acid do you require right so all you need to do is simply to apply the formula c1 times v1 equals c2 times v2 all right and here's the answer the concentration of the stock solution is 12 m and v1 is the volume that you need to find out the volume of the concentrated acid that you require the final concentration of the diluted solution is 22 m and it's volume is 1.5 liters and from here you can calculate a v1 to be 25 milliliters okay next example a 500 ml solution of point to M Koh is prepared by diluting 50 ml of the stop solution or the concentrated solution calculate the molarity or calculate the concentration of the stock solution again you apply the formula c1v1 equals c2v2 and here's the answer well so what you want is the concentration of the stop solution right which is c1 the volume of the stock solution is 50 ml in liters is point zero five liters and the concentration of the diluted solution is point two M and the volume is point five liters from here you can calculate out the molarity the concentration of the stop solution to be two and from here we move on to another important phrase that you need to know is call dilution factor let's look at a simple example here let's say or regionally you have 20 ml of stop and then you add some solvent right and form a 50ml diluted solution right so the stock solution we say has been diluted 2.5 times right from 20 to 50 so the stock solution has been diluted 2.5 times so we say that the dilution factor is 2.5 the dilution factor can be calculated in these two different ways the dilution factor equals the volume of the final diluted solution divided by the volume of the initial concentrated solution or the dilution factor can be calculated this way valuation factor equals the concentration of the initial concentrated solution divided by the concentration of the final diluted solution all right so these two values are exactly the same and you know the dilution factor is always greater than 1 so using this two expression we go on to the next example from a stop solution of 0.5 M sodium chloride you want to obtain 100 ml of a solution this diluted 50 for what volume of the stop solution do you require huh so this is the simple answer you know that the final volume is 100 ml right this is the volume of the diluted solution what you want to find out is the initial initial volume of the stock solution that you need you know that the dilution factor is 50 and then you recall the dilution factor can be expressed this way as volume of the final diluted solution divided by volume of the initial concentrated solution so from this expression you make volume of the initial concentration solution to be the subject so the volume of the initial concentration are concentrated solution equals V final divided by dilution factor which is 100 divided by 50 gives you 2 ml so you need 2 ml of the spot solution to prepare the diluted solution right so please take note the concentration of sodium chloride solution does not play a part in the calculation here