A Note on Dilution and Concentrations of Solutions
Problems dealing with dilution of solutions from one concentration to another concentration can sometimes be confusing.
In order to have a better understanding of it, we will use graphical representation of the solute-solvent components of a solution to help us visualize the problem.
A typical example of this kind of problem is given below:
Given 2 L of 2 M NaCl solution, make a solution having a concentration of 1 M NaCl solution. What must be the final volume of the solution?
We will use a graphical representation like the one below to explain this problem.
The blue box represents one liter of solution, and each white ellipse represents one mole of solute ( NaCl ).
The blue box containing the two ellipses represents the concentration of the solution ( 2 M NaCl = 2 moles NaCl/1 liter of solution ).
So our given solution of 2 L of 2 M NaCl will be represented like this:
By inspection, we can see that we have a total of 4 moles NaCl in 2 L of solution.
Mathematically, finding the total amount of solute in a solution is equivalent to:
Volume of solution (V) x Concentration of solution (C) = amount of solute
or simply,
V x C = amount of solute (1)
Since the problem requires us to make 1 M NaCl solution from the given solution by diluting it (adding water to the solution), the total amount of solute must remain the same before and after dilution.
To account for the 4 moles of NaCl, the graphical representation for our final solution will be like this:
By inspection again, we can see that the volume of the final solution is now increased to 4 liters.
Using equation number 1, we can find the final volume of the solution mathematically by solving for V:
So the problem can be represented like this:
| 2 L of 2 M NaCl | 4 L of 1 M NaCl | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
| amount of solute before dilution | = | amount of solute after dilution | |||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
Substituting equation 1 into this, and using subscript 1 and 2 to distinguish the initial solution from the final solution, gives us:
V1 x C1 | = | V2 x C2 (2) |
Equation 2 is the formula for dilution problems involving different kinds of concentrations.
Solving for V2:
Just knowing that the amount of solute does not change and employing the dimensional analysis to the problem will enable you to figure out the formula by yourself, as long as you know the definitions of different concentrations.
Here are the commonly used concentrations in chemistry problems and their definitions:
The last two expressions are used in very dilute solutions. They are used to express trace amount of solute in aqueous solutions. (1 gram = 1,000 milligrams = 1,000,000 micrograms)
Problems
The best way to improve your problem solving skill in chemistry is to solve as many chemistry problems as you can find. Here are some problems on dilutions and concentrations of solutions. The answer is provided for each problem.
Dilution Problems
| 1. | What is the volume of the diluted solution of 10.0 mL of 4.34 M K2HPO4 if its final concentration is 0.55 M K2HPO4? |
| Answer: 78.9 mL | |
| 2. | If 15.0 mL of 1.9 M Cu(NO3)2 solution is diluted to 45.0 mL, calculate the new molarity of the solution. |
| Answer: 0.6 M | |
| 3. | 23.6 mL of 1.07 M K2SO4 solution was prepared from an unknown volume of 4.21 M K2SO4 solution. What was the initial volume of the solution? |
| Answer: 6.0 mL | |
| 4. | 20.00 mL of an unknown concentration of LiNO3 solution was diluted to 64 mL to make a concentration of 0.59 M LiNO3. Determine the initial concentration of the diluted solution. |
| Answer: 1.89 M | |
| 5. | 22.00 mL of an unknown concentration of NaClO3 solution was diluted to 63 mL to make a concentration of 0.5 M NaClO3. Determine the initial concentration of the diluted solution. |
| Answer: 1.43 M | |
| 6. | If 8.0 mL of 0.8 M NiCl2 solution is diluted to 63.0 mL, calculate the new molarity of the solution. |
| Answer: 0.1 M | |
| 7. | 20.4 mL of 1.44 M NH4HCO3 solution was prepared from an unknown volume of 1.63 M NH4HCO3 solution. What was the initial volume of the solution? |
| Answer: 18.0 mL | |
| 8. | 21.00 mL of an unknown concentration of KCH3COO solution was diluted to 64 mL to make a concentration of 0.3 M KCH3COO. Determine the initial concentration of the diluted solution. |
| Answer: 0.9 M | |
| 9. | What is the volume of the diluted solution of 17.0 mL of 4.06 M NaCl if its final concentration is 2.84 M NaCl? |
| Answer: 24.3 mL | |
| 10. | If 24.0 mL of 3.9 M K2Cr2O7 solution is diluted to 38.0 mL, calculate the new molarity of the solution. |
| Answer: 2.5 M | |
Concentration Problems
| 1. | Determine the weight in grams of the solute contained in a given volume and a given concentration of the following solutions: (a) 3.0 L of 1.97 M Ca(NO3)2.4H2O solution, (b) 14.0 mL of 1.84 M CuS solution, (c) 23.0 mL of 0.91 M Cs2SO4 solution. (mol. wt.: Ca(NO3)2.4H2O=236.1506, CuS=95.606, Cs2SO4=361.8684) |
| Answer: (a) 1395.7 g, (b) 2.5 g, (c) 7.6 g | |
| 2. | Calculate the final molarity (M) of a mixed solution of 15.0 mL of 4.65 M NaCH3COO solution and 10.0 mL of 2.96 M NaCH3COO solution, assuming their volumes are additive. |
| Answer: 4 M | |
| 3. | Calculate the resulting molarity (M) of NO3-1 from the mixture of 17.00 mL of 1.42 M NaNO3 solution and 18.00 mL of 4.2 M Mg(NO3)2 solution, assuming their volumes are additive. |
| Answer: 5.01 M | |
| 4. | Determine the weight in grams of the solute contained in the following solutions: (a) 14.0 mL of 0.78 M KOH solution, (b) 4.0 L of 1.81 M Ca(CH3COO)2 solution, (c) 6.0 L of 3.03 M AgNO3 solution. (mol. wt.: KOH=56.1056, Ca(CH3COO)2=158.169, AgNO3=169.8729) |
| Answer: (a) 0.6 g, (b) 1145.1 g, (c) 3088.3 g | |
| 5. | What is the molarity (M) of the following solutions: (a) 96% (w/w) glacial acetic acid solution with a density of 1.06 g/mL, (b) 35% (w/w) ammonium hydroxide solution with a density of 0.88 g/mL. (mol. wt.: CH3COOH=60.0524, NH4OH=35.0456) |
| Answer: (a) 16.95 M, (b) 8.79 M | |
| 6. | Calculate the resulting molarity (M) of Cl-1 from the mixture of 10.00 mL of 2.25 M CoCl3 solution and 20.00 mL of 3.2 M MgCl2 solution, assuming their volumes are additive. |
| Answer: 6.52 M | |
| 7. | Calculate the weight of the solute required in order to prepare the following solutions: (a) 18.0 mL of 5.21 M Na4P2O7 solution, (b) 2.0 L of 1.7 M NH4H2PO4 solution, (c) 5.0 L of 1.82 M Al2(SO4)3 solution. (mol. wt.: Na4P2O7=265.9024, NH4H2PO4=115.02546, Al2(SO4)3=342.13588) |
| Answer: (a) 24.9 g, (b) 391.1 g, (c) 3113.4 g | |
| 8. | What is the molarity (M) of the following solutions: (a) 40% (w/w) hydrobromic acid solution with a density of 1.38 g/mL, (b) 97% (w/w) concentrated sulfuric acid solution with a density of 1.84 g/mL. (mol. wt.: HBr=80.9119, H2SO4=98.0734) |
| Answer: (a) 6.82 M, (b) 18.2 M | |
| 9. | Determine the weight in grams of the solute contained in the following solutions: (a) 9.0 mL of 1.4 M NaClO3 solution, (b) 1.0 L of 4.24 M Zn(NO3)2.6H2O solution, (c) 4.0 L of 3.31 M Ca(NO3)2.4H2O solution. (mol. wt.: NaClO3=106.44097, Zn(NO3)2.6H2O=297.481, Ca(NO3)2.4H2O=236.1506) |
| Answer: (a) 1.3 g, (b) 1261.3 g, (c) 3126.6 g | |
| 10. | What is the molarity of a 23.0 mL solution if it contains: (a) 6.23 g of CdCO3, (b) 1.78 g of K2S2O5, (c) 7.47 g of Na2HPO4.2H2O? (mol. wt.: CdCO3=172.4192, K2S2O5=222.3136, Na2HPO4.2H2O=177.9892) |
| Answer: (a) 1.6 M, (b) 0.3 M, (c) 1.8 M | |
