To carry out a moles concentration calculation, follow these steps:

1. Determine the amount of solute in moles: This can be done by dividing the mass of the solute by its molar mass, which is the sum of the atomic masses of all the atoms in the compound. For example, if you have 10 grams of sodium chloride (NaCl), which has a molar mass of 58.44 g/mol, the amount of NaCl in moles would be:

10 g / 58.44 g/mol = 0.171 moles of NaCl

2. Measure the volume of the solution in dm³ (litres): Use a graduated cylinder or volumetric flask to accurately measure the volume of the solution as precise as possible.
3. Calculate the concentration (molarity) of the solution: Use the concentration equation to calculate the molarity of the solution. For example, if you have 250 mL of a solution containing 0.171 moles of NaCl, the molarity of the solution would be:

Volume = 250ml/1000 = 0.25 dm³

Molarity (M) = 0.171 mol / 0.25 dm³ = 0.684 mol dm^-3

In this case, the concentration of the solution is 0.684 mol dm^-3 or 0.684 M.

It is important to note that when carrying out moles concentration calculations, it is essential to use the correct units for both the amount of solute and the volume of the solution to ensure accuracy. Also, ensure that the molar mass of the solute is calculated correctly to avoid errors in the calculation.

Titrations

In a titration practical, the concentration of an unknown solution can be determined by titrating it against a known solution of known concentration. The volume of the known solution required to completely react with the unknown solution is measured, and from this information, the concentration of the unknown solution can be calculated using mole calculations.

Here is a worked example:

A student wants to determine the concentration of a hydrochloric acid (HCl) solution by titrating it against a sodium hydroxide (NaOH) solution of known concentration. The student adds 25.0 mL HCl to a flask, and slowly titrates it with 0.15 M NaOH until the reaction is complete. The volume of NaOH required to reach the endpoint is 30.0 mL. What is the concentration of the HCl solution?

To solve this problem, we first need to write a balanced equation for the reaction:

HCl + NaOH → NaCl + H₂O

We know that the reaction is a 1:1 stoichiometry, meaning that for every mole of HCl, we need one mole of NaOH to completely react.

Using the volume and concentration of NaOH, we can calculate the number of moles of NaOH used in the reaction:

moles of NaOH = concentration x volume moles of NaOH = 0.15 mol dm^–³ x 0.030 dm³ moles of NaOH = 0.0045 moles

Since the reaction is a 1:1 stoichiometry, the number of moles of HCl used in the reaction is also 0.0045 moles.

Now, we can use the number of moles of HCl and the volume of the HCl solution to calculate its concentration:

concentration of HCl = moles of HCl / volume of HCl solution concentration of HCl = 0.0045 moles / 0.025 dm³ concentration of HCl = 0.18 mol dm^–³

Therefore, the concentration of the HCl solution is 0.18 mol dm^–³.

It is important to note that when carrying out titration practicals, accuracy in measuring the volume of the solutions used is crucial to the success of the experiment. Any errors in the volume measurement can lead to inaccurate results. Also, it is essential to ensure that the reaction has reached its endpoint before reading the final volume of the titrant.

Questions:

1. What volume of 0.10 M HCl is needed to neutralise 25.0 mL of 0.20 mol dm^-3 NaOH solution?
2. How many moles of NaOH are in 50.0 mL of a 0.15 mol dm^-3 NaOH solution?
3. What volume of 0.20 mol dm^-3 HCl is needed to neutralise 0.050 moles of Ca(OH)₂?
4. If 25.0 mL of 0.10 mol dm^-3 H₂SO₄ is added to 75.0 mL of 0.15 mol dm^-3 NaOH, what is the concentration of the resulting solution?
5. What is the molarity of a solution if 0.025 moles of solute are dissolved in 500.0 mL of solution?
6. What mass of Mg(OH)₂ is needed to neutralise 50.0 mL of 0.25 mol dm^-3 HCl solution?
7. What volume of 0.10 mol dm^-3 LiOH is needed to titrate 10.0 mL of 0.15 mol dm^-3 H₃PO₄ solution?
8. How many moles of HCl are needed to neutralise 0.050 moles of RbOH?
9. If 50.0 mL of 0.10 mol dm^-3 NaOH is added to 75.0 mL of 0.15 mol dm^-3 H₂SO₄, what is the concentration of the resulting solution?
10. What volume of 0.20 mol dm^-3 KOH is needed to neutralise 0.100 moles of H₃PO₄?

Answers:

1. 50 mL HCl
2. 0.0075 moles
3. 500.0 mL
4. 0.121 mol dm^-3
5. 0.050 mol dm^-3
6. 0.01875 g
7. 15.0 mL
8. 0.050 moles
9. 0.128 mol dm^-3
10. 25.0 mL

• Practical Chemistry