Calculate the new molarity if each of the following...

Calculate the new molarity if each of the following dilutions is made. Assume the volumes are additive. a. $55.0 \mathrm{~mL}$ of water is added to $25.0 \mathrm{~mL}$ of $0.119 \mathrm{M} \mathrm{NaCl}$ solution b. $125 \mathrm{~mL}$ of water is added to $45.3 \mathrm{~mL}$ of $0.701 \mathrm{M} \mathrm{NaOH}$ solution c. $550 . \mathrm{mL}$ of water is added to $125 \mathrm{~mL}$ of $3.01 \mathrm{M}$ KOH solution d. $335 \mathrm{~mL}$ of water is added to $75.3 \mathrm{~mL}$ of $2.07 \mathrm{M} \mathrm{CaCl}_{2}$ solution

Calculate the new molarity if each of the following dilutions is made. Assume the volumes are additive.
a. $55.0 \mathrm{~mL}$ of water is added to $25.0 \mathrm{~mL}$ of $0.119 \mathrm{M} \mathrm{NaCl}$ solution
b. $125 \mathrm{~mL}$ of water is added to $45.3 \mathrm{~mL}$ of $0.701 \mathrm{M} \mathrm{NaOH}$ solution
c. $550 . \mathrm{mL}$ of water is added to $125 \mathrm{~mL}$ of $3.01 \mathrm{M}$ KOH solution
d. $335 \mathrm{~mL}$ of water is added to $75.3 \mathrm{~mL}$ of $2.07 \mathrm{M} \mathrm{CaCl}_{2}$ solution

Key Concepts

Molarity

Molarity is a measure of the concentration of a solution, expressed as the number of moles of solute per liter of solution. It is a fundamental concept in chemistry for quantifying the concentration of a chemical species in a solution, which is crucial for reaction stoichiometry, titrations, and various other analytical techniques.

Dilution

Dilution refers to the process of reducing the concentration of a solute in a solution by adding more solvent. The key principle in dilution is that the number of moles of solute remains constant; only the volume of the solution changes. This concept is typically applied using the relationship M1 × V1 = M2 × V2, where M and V represent molarity and volume, respectively.

Volume Additivity

Volume additivity is the assumption that the total volume of a solution after mixing two liquids is equal to the sum of their individual volumes. This concept simplifies calculations in dilution problems by allowing one to add the volume of the solute solution to the volume of the added water or solvent to obtain the final volume needed for concentration calculations.

Conservation of Solute

The conservation of solute principle states that during a dilution process, the amount of solute (in moles) does not change even though the solution’s volume increases. This key concept underlies the dilution equation, ensuring that the product of the initial molarity and volume equals that of the final molarity and volume.

Transcript

00:01 So when you have a dilution, the number of moles stays the same before and after.

00:05 So the first molarity times volume is equal to the second molarity times volume.

00:10 So if you have to calculate the new molarity upon a dilution, we'll just solve this formula for m2 by dividing both sides by v2.

00:18 So you have m1 v1 over v2.

00:24 So if you have 55 milliliters of water added to 25 milliliters of 0 .119 molar solution, that means that your initial molarity is, this is for part a, 0 .119 molar, and then you have the initial volume is 25 milliliters, and then you divide by your final volume, which is the sum, the 25 plus the 55 that you've added, which is going to be 80 milliliters.

01:10 And that will give you m2, which is equal to, when you go on your calculator, 0 .119 times 25 divided by 80, you get 0 .037 using the right number of significant figures, molar.

01:49 Then if we have, for part b, we have 45 .3 milliliters of a 0 .701 .1 .1.

02:03 Molar solution.

02:05 So we'll put the initial molarity times the volume, and then we add 125 milliliters.

02:23 So that means that our final volume is going to be 125 plus 45 .3, which is 170 .3 milliliters.

02:37 And that's equal to m2, the second molarity, which is going to be 0 .701 times 45 .3, divided by 170 .3, which equals 0 .186.

03:11 That made clear part a and part b to make room for c and d.

03:15 It's the same formula, so i'll leave that up there...

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