Aqueous Solutions - Molarity An aqueous solution consists of at least two components, the solvent (water) and the solute (the stuff dissolved in the water). Usually one wants to keep track of the amount of the solute dissolved in the solution. We call this the concentrations. One could do by keeping track of the concentration by determining the mass of each component, but it is usually easier to measure liquids by volume instead of mass. To do this measure called molarity is commonly used. Molarity (M) is defined as the number of moles of solute (n) divided by the volume (V) of the solution in liters.
It is important to note that the molarity is defined as moles of solute per liter of solution, not moles of solute per liter of solvent. This is because when you add a substance, perhaps a salt, to some volume of water, the volume of the resulting solution will be different than the original volume in some unpredictable way. To get around this problem chemists commonly make up their solutions in volumetric flasks. These are flasks that have a long neck with an etched line indicating the volume. The solute (perhaps a salt) is added to the flask first and then water is added until the solution reaches the mark. The flasks have very good calibration so volumes are commonly known to at least four significant figures. Example #1 : Molarity Calculation The equation for calculating Molarity from the moles and volume is very simple. Just divide moles of solute by volume of solution.
What is the molarity (with the correct numbers of significant figures) of a 0.40 moles of NaCl dissolved in 0.250 liters? Answer Example #2 : Making Dilutions
The volume units must be the same for both volumes in this equation. In general, M1 usually refers to as the initial molarity of the solution. V1 refers to the volume that is being transferred. M2 refers to the final concentration of the solution and V2 is the final total volume of the solution. Remeber that the number of moles of solute does not change when more solvent is added to the solution. Concentration, however, does change with the added amount of solvent. (illustration) Don't forget this concept. You will use it again in acid-base equilibrium. Dilution calculation example: How do you prepare 100ml of 0.40M MgSO4 from a stock solution of 2.0M MgSO4? Answer: There are two solutions involved in this problem. Notice that you are given two concentrations, but only one volume. Solution #1 is the one for which you have only concentration - the solution that is already sitting on the shelf. Solution #2 is the one for which you have both concentration and volume - the solution that you are going to prepare. At least until you are comfortable with this type of problem, it may be helpful to write out what numbers go with what letters in our equation. M1 = 2.0M MgSO4 ; V1 = unknown
In order to continue enjoying our site, we ask that you confirm your identity as a human. Thank you very much for your cooperation.
Learning Objectives
Up to this point we have used stoichiometry to "count" atoms, molecules and ions by measuring mass of pure substances and using molar masses to calculate the number of chemical entities (moles). Many of the reactions we have studied involve solutions, where we are interested in a solute that is dissolved in a solvent, and we can not measure the mass of the solute independent of the solvent. We also need to realize that many chemical reactions require the reactants to be mobile, where they can bump into each other, and this can occur when they are dissolved in a solvent. So it is very important that we can count chemical entities when the entity of interest is a solute dissolved in a solvent, and in which case we measure the mass of volume of the solution as a whole (solvent plus all solutes) of which volume is typically the easiest to measure.
There are two basic ways of reporting the concentration of a solute in a solvent, by reporting the mass of solute in a given volume, or the number of moles of solute in a given volume. These are effectively conversion factors that define the equivalent mass or moles of a solute to the volume of the solution.
So a 3.0M solution of sucrose has 3.0 mole of sucrose per liter Figure \(\PageIndex{1}\): In order to avoid confusion we will use a small "m" to denote mass, and a large ""m" to denote Molarity.How do we convert between mass and mole based concentration? Simply by multiplying or dividing by the molar mass (g/mol). If you know the g/L, you simply divide by the molar mass, and if you know the moles/L and want the mass/L, you multiply by the molar mass (just use the molar mass as an equivalence statement). How do we "count" solute molecules and ions? By measuring the volume and knowing the molarity. This is analagous to measuring the mass and knowing the molar mass for a pure substance. \[n(moles)=M(\frac{mol}{L})(V(L)) \\ n=MV\] This can be contrasted to how we "count" chemical entities that are pure solids. \[n(moles)=\frac{m(g)}{molar \; mass (\frac{g}{mol})} \\ n=\frac{m}{fw}\]
Step 1: Calculate Mass of solute needed for desired volume Step 2: Quantitatively Transfer Mass to Volumetric Flask Step 3: Dilute to volume with solvent, ensuring that all of the solid has dissolved.
Making 500.0 mL of 0.500 M Copper(II)Sulfate Step 1: Calculate Mass CuSO4(s) needed. \[0.5000L\left (\frac{0.500molCuSO_{4}}{L} \right )\left ( \frac{159.6gCuSO_{4}}{mol} \right )=39.9gCuSO_{4}\] Step 2: Weight 39.9g CuSO4(s) and quantitatively transfer to 500 mL graduated cylinder (which is calibrated to 500.0 mL), being sure all the salt is transferred. Step 3: Fill half way and mix, then dilute to volume. Make sure all the solute is dissolved, and recheck that solution level is at bottom of meniscus.
Exercise \(\PageIndex{1}\) What is the molarity of a solution made when 66.2 g of C6H12O6 are dissolved to make 235 mL of solution? Answer1.57 M C6H12O6 Why can't you just add 500.0mL of water to 39.9 g CuSO4 to make 500.0 mL of 0.500M solution?
Exercise \(\PageIndex{2}\) Using concentration as a conversion factor, perform the following calculation. A) How many liters of 0.0444 M CH2O are needed to obtain 0.0773 mol of CH2O? B) What mass of solute is present in 1.08 L of 0.0578 M H2SO4? C) What volume of 1.50 M HCl solution contains 10.0 g of hydrogen chloride? AnswerA) 1.74 L B) 6.12 g C) 183 mL or 0.183L
When an ionic compound dissolves it breaks up into its ions. (NH4)2Cr2O7(aq) --> 2NH4+(aq) + Cr2O7(aq) Dissolving 252.07 grams of ammonium dichromate (fw = 252.07 g/mol) results in a solution that is 1M ammonium dichromate, but when an ionic salt dissolves, it breaks up into ions, and so what you really have is a solution that is 2 M ammonium ion (NH4+) and 1 M in dichromate ion (Cr2O7+2)  Dissolution of 1 mol of an Ionic Compound. In this case, dissolving 1 mol of (NH4)2Cr2O7 produces a solution that contains 1 mol of Cr2O72− ions and 2 mol of NH4+ ions. (Water molecules are omitted from a molecular view of the solution for clarity.)
Example \(\PageIndex{1}\) Solution Concentration of Iron(II)chloride Water is added to 2.16 g of the ionic compound ferrous chloride to make a solution with a total volume of 100.0 mL. Express the concentration of the salt solution, and that of its ions. 1. What is the salt concentration? Solution Video \(\PageIndex{1}\) goes over this calculation. Ans: 0.170M FeCl2(aq) 0.170M Fe+2(aq) 0.340M Cl-(aq) Yes, there are three answers! (that of the salt, and of the ions) Video \(\PageIndex{1}\): 3'36" YouTube calculating the concentration of a Ferrous Chloride in example \(\PageIndex{1}\).
Exercise \(\PageIndex{3}\) You have a 1.50 M solution of Na2CO3. What is the concentration of: A) sodium ions? B) carbonate ions? C) total ions? AnswerA) [Na+] = 3.00 M B) [CO3-2]= 1.50 M C) [Na+] + [CO3 -2] = 4.50 M ; or you can choose think for every 1 mole of Na2CO3, there are 3 moles of ions (identify of the ions is irrelevant).
Many stockroom reagents come as concentrated solutions that can easily be diluted to a desired concentration by adding solvent. Adding a solvent does not change the moles of solute (n), so
Step 1: Calculate initial volume of stock Step 2: Transfer to volumetric flask with a volumetric pipette Step 3: Dilute to Volume
Example \(\PageIndex{2}\) How do you make 500.0 ml of 0.70M HCl solution from stock 11.6 M HCl? Solution Step 1: Calculate initial volume of 11.6 M hydrochloric acid needed. \[M_{i}V_{i}=M_{F}V_{F}\] Step 2: Quantitatively transfer this volume to a 500 mL volumetric flask (this has 4 significant Figures, that is, it is calibrated to 500.0 mL) Step 3: Dilute with water to mark Video \(\PageIndex{2}\): Dilution problem on creating 500.0 ml of a 0.700M HCl solutioin
Exercise \(\PageIndex{4}\) What is the concentration of an HCl solution if 598 mL of 0.778 M HCl is diluted to 1.00 L? Answere. 5.23 * 102 mL \(M_{i}*V_{i}=M_{f}*V_{f}\) \(M_{f}=\frac{\left (M_{i}*V_{i} \right )}{M_{f}}=\frac{ 0.778 M\left ( 0.598L \right )}{1.00 L}= 0.465 M \)
Robert E. Belford (University of Arkansas Little Rock; Department of Chemistry). The breadth, depth and veracity of this work is the responsibility of Robert E. Belford, . You should contact him if you have any concerns. This material has both original contributions, and content built upon prior contributions of the LibreTexts Community and other resources, including but not limited to:
|
What are the 3 ways to calculate concentration?
Pos Terkait
Copyright © 2024 membukakan Inc.
