In chemistry, understanding the concept of molarity is important for calculating the amount of solute present in a solution. Molarity (M) is defined as the number of moles of solute per liter of solution. Knowing the molarity and the volume allows us to determine the number of moles of solute present.
Summary
For a 0.10 M CaCl2 solution with a volume of 250 mL, the number of moles of CaCl2 can be calculated as follows:
- Molarity (M) = moles of solute (n) / liters of solution (L)
- The given molarity is 0.10 M CaCl2
- The given volume is 250 mL = 0.250 L
- Use the molarity equation to find the moles of CaCl2:
- n = (M)(L) = (0.10 M)(0.250 L) = 0.025 moles CaCl2
Therefore, the number of moles of solute (CaCl2) present in 250 mL of a 0.10 M solution is 0.025 moles.
Step-by-Step Solution
Let’s go through the full step-by-step process of using the molarity equation to find the moles of solute:
- Write the given information:
- Molarity of CaCl2 solution = 0.10 M
- Volume of CaCl2 solution = 250 mL
- Convert volume into liters:
- 250 mL x (1 L / 1000 mL) = 0.250 L
- Write the molarity equation:
- Molarity (M) = moles of solute (n) / liters of solution (L)
- Rearrange the equation to solve for moles:
- n = (M)(L)
- Plug in the known values:
- n = (0.10 M)(0.250 L)
- Calculate:
- n = 0.025 moles CaCl2
So for a 0.10 M CaCl2 solution with a volume of 250 mL, there are 0.025 moles of CaCl2 present.
Explanation of Calculations
Let’s break this down step-by-step to understand what these calculations represent:
- Molarity (M) is measured in moles of solute per liter of solution. It is a concentration unit that tells us the number of moles of solute present in 1 L of solution.
- The molarity given in this problem is 0.10 M. This tells us there are 0.10 moles of CaCl2 present in every 1 L of the solution.
- The volume given is 250 mL. We converted this to liters to match the units of molarity. 250 mL is equal to 0.250 L.
- Using the molarity equation, we calculated the number of moles by multiplying the molarity (0.10 M) by the volume in liters (0.250 L).
- This gave us 0.025 moles of CaCl2 present in 0.250 L of solution, or 250 mL of solution.
So the molarity equation allowed us to use the concentration (molarity) and volume of solution to find the moles of solute. This is a very useful tool in chemistry calculations.
Common Mistakes
Here are some common mistakes when calculating moles from molarity and volume:
- Forgetting to convert volume into liters before plugging into the equation
- Plugging volume in mL directly into the molarity equation instead of converting to L
- Calculating molarity incorrectly from moles and volume
- Using the wrong units throughout the calculation
- Not paying attention to significant figures and rounding too early
To avoid these errors, be sure to:
- Always convert volume to liters before using in molarity equation
- Use units of moles, liters, and molarity consistently
- Round only at the end of the calculation to preserve significant figures
- Double check that your units cancel properly in the molarity equation
Practice Examples
Let’s do a few practice examples calculating moles from molarity and volume:
Example 1
Molarity of KOH solution = 0.150 M
Volume of solution = 325 mL
- Convert volume: 325 mL x (1 L / 1000 mL) = 0.325 L
- Use molarity equation: n = (M)(L)
- Plug in values: n = (0.150 M)(0.325 L) = 0.0488 moles KOH
Example 2
Volume of HNO3 solution = 135 mL
Moles of HNO3 in solution = 0.0180 moles
- Convert volume: 135 mL x (1 L / 1000 mL) = 0.135 L
- Rearrange equation to find molarity: M = n/L
- Plug in values: M = 0.0180 moles / 0.135 L = 0.133 M
Be sure to practice converting between moles, volume, molarity, and mass of solute when solving problems. Pay close attention to units throughout the calculations.
Solubility Rules
When dealing with aqueous solutions, it’s important to know if a solute will dissolve in water or not. The solubility rules provide guidelines for the solubility of ionic solids in water based on the ions present.
Some common solubility rules are:
- All sodium, potassium, and ammonium salts are soluble.
- Nitrates, acetates and chlorides are generally soluble.
- Sulfates are soluble, except calcium, barium, lead, and silver sulfates.
- Carbonates, phosphates, sulfides, and oxides are generally insoluble, except for compounds of the Group 1A and ammonium ions.
Knowing these solubility rules allows you to determine if a given ionic compound will dissolve in water to form an aqueous solution. This is important when calculating the molarity or moles of solute in a solution.
Molarity and Solution Preparation
Understanding molarity also allows chemists to prepare solutions of a desired concentration. For example, say you need 250 mL of a 0.15 M NaCl solution for an experiment:
- Calculate moles of solute needed:
- n = (M)(L)
- n = (0.15 M)(0.250 L) = 0.0375 moles NaCl
- Find the molar mass of NaCl from periodic table:
- NaCl molar mass = 58.44 g/mol
- Convert moles NaCl to mass:
- m = n x molar mass
- m = 0.0375 moles NaCl x 58.44 g/mol = 2.19 g NaCl
- Dissolve 2.19 g NaCl in water and add water until final volume is 250 mL. This gives a 0.15 M NaCl solution.
This demonstrates how you can use molarity to calculate the precise mass of solute needed to prepare a solution of the desired concentration and volume.
Dilution of Solutions
Another useful application of molarity is diluting concentrated solutions to reach a lower desired concentration. The dilution equation is:
- M1V1 = M2V2
Where:
- M1 = initial molarity
- V1 = initial volume
- M2 = final molarity after dilution
- V2 = final volume after dilution
An example of diluting 100 mL of 6 M HCl to make a 0.10 M HCl solution with a final volume of 1 L:
- M1 = 6 M
- V1 = 0.100 L
- M2 = 0.10 M
- V2 = 1 L
- Plug into dilution equation: (6 M)(0.100 L) = (0.10 M)(1 L)
- Solve for V1: V1 = 16.7 mL of 6 M HCl should be diluted to 1 L to make a 0.10 M solution.
Dilution calculations are very useful in the chemistry lab for preparing solutions of specific molarities.
Titration Calculations
Molarity also comes into play when performing titrations in the laboratory. A titration involves slowly adding a solution of known concentration (the titrant) to another solution of unknown concentration (the analyte) until the reaction is complete.
For example, say we need to find the molarity of an HCl solution. We can titrate it against a 0.100 M NaOH solution:
- HCl (analyte) + NaOH (titrant) → NaCl + H2O
It takes 25.00 mL of the 0.100 M NaOH to completely react with the HCl. We can use the molarity equation to find the moles of NaOH added:
- n = (M)(L)
- n = (0.100 M)(0.02500 L) = 0.00250 moles NaOH
Then we use the stoichiometric ratio between HCl and NaOH (1:1) to find the moles and molarity of the HCl:
- moles HCl = moles NaOH = 0.00250 moles
- Molarity = moles / liters
- Molarity of HCl = 0.00250 moles / 0.02500 L = 0.100 M
Using titration and stoichiometry, the molarity of unknown solutions can be experimentally determined.
pH and Molarity
For aqueous solutions of strong acids and bases, the molarity is directly related to the pH of the solution. The Henderson-Hasselbalch equation describes this relationship:
- pH = pKa + log(acid molarity / base molarity)
Where pKa is dependent on the identity of the acid/base.
For example, for a 0.10 M acetic acid solution:
- pKa (acetic acid) = -log(1.8 x 10-5) = 4.74
- pH = 4.74 + log(0.10/x) (x = molarity of A-)
- x can be approximated as the molarity for a weak acid
- pH = 4.74 + log(0.10/0.10) = 4.74
Therefore, a 0.10 M solution of acetic acid will have an approximate pH of 4.74.
The molarity and pKa of an acid/base allow the pH to be easily calculated.
Conclusion
In summary, molarity is an extremely useful concentration unit in chemistry. By understanding the concept of molarity, one can determine the moles of solute in a solution from the molarity and volume. Molarity also allows chemists to prepare solutions of a desired concentration, dilute solutions, perform titrations, and calculate pH. Learning to calculate using molarity and manipulate this unit will greatly aid chemistry students in both classroom and laboratory applications.
