4.4 Molarity and Dilutions Flashcards
Molarity helps meausre concentration
Molarity = Moles of solute/Liters of solution
Keep in mind this is a measure of concentration
Solution = Solvent + Solute
n (number of moles) M = -----
VL (volume in liters of solution)
If you have any 2 of those variables you can solve for it algebrically
By re arranging we get: m * VL = n
* This is important because its a new way we can calculate moles
* basically it gives us the # of moles of solute of that solution
* because once you know the moles you can convert that to grams, molecules/atoms etc…
example below
What volume of 5M (molar = molarity) NH3 contains 0.10 moles of NH3
Dilutions
This is taking a solution from being more concetrated in a certain solute to being less concentrated
* typically accomplushed by adding more solvent
Dilution equations:
m1v1 = m2v2
c1v1 = c2v2
Lets see I like coolaid, and i like it a lot. So instead of putting 1 packet into a pitcher, I put 30 in. I taste it and I can taste that its wayyyy to concentrated, so I need to dilute it. So What i do is poor that pitcher w/ really concnetrated coolaid into my bathtub, and i fill up my bathtub. I then take a glass full of the new centrated coolaid and its just right –> because its been diluted to the better conetration
* So 30 packets were in the original container, meaning 30 ended up in the bathtub
* So the amount of solute going from the pitcher to the bathtub didnt change, the amount of solvent did.
* So typically in these calculations the # of moles of solute that you start w/ is the same as the # of moles of solute you end w/ its just the amount of solvent that changes to end up w/ more overall solution
From our molarity equation we can see that the moles of solute = Molarity * V (in L)
* And thats where this first equation comes from (M1V1 = M2V2)
* Molarity1 * L1 = Molarity2 * L2
* You’re just saying the inital # of moles of solute = the final # of moles of solute
* becuase remember moles of solute = Molarity * V (in L), which is the same as that equation
* its saying “inital # of moles of solute = final number of moles of solute”
* which is true, think back to our bathtub example, its the solvent thats changing not the solute
* For the new equation above we can use mL or any volume measurement, as long as we keep it the same on both sides because they will cancel out - we can honestly use any concentration unit because it will cancel out
* Because of the point above we can now use the equation C1V1 = C2V2 - where c stands for any concentration - key is that we have to use the same # for concetration on either side
To what final volume must 20.0mL of 0.40M NaCl be diluted to result in a 0.080M solution
If it was asking how much water was added, then it would be the difference between V2 and V1. However, its simply asking for the final volume
Molar concentration = molarity
Have to memorize the equation for molarity = mol solute / L solution
* needs to be in L
What a 0.5M solution of NaCl means
Molarity = Molues solute/L solution
In this case it would mean that there are 0.5M of NaCl per liter of solution
0.5 mol means
* 0.5 x 6.022x10^23 = 3.011 x 10^23 formula units
* So in a 0.5 MNaCl solution, every liter contains 3.011 x 10^23 NaCl formula units in dissolved water
So if I have 0.08M NaOH, thats the same as saying I have 0.08 moles of NaOH per 1L of solution.
* so thats why molarity can also be called concentration. This is the starting concnetration in this problem. However, my goal is to find the final concentraiton, which is molarity2
Key was seeing that I needed to calculate how much needed to be added. Not just the final concentration. So you would need to subtract the starting volume from the ending one
So its essentially saying that we need to dilunte the solution to 900mL to get the concentration of KBr to go down to 0.1M (0.1mol KBr for every 1L of solution)
