Understanding Concentration in Chemistry: Its Calculation and Connection to Titrations
Keywords:
concentration, molarity, moles, solution, titrations, acid-base reactions, stoichiometry
Introduction:
Concentration is a fundamental concept in chemistry that indicates the amount of a substance (solute) dissolved in a given volume of solution. It plays a vital role in various chemical processes, including titrations. In this blog post, we will delve into the concept of concentration, how it is calculated, and its relationship with titrations.
What is Concentration? In chemistry, concentration refers to the measure of how much solute is present in a given volume of a solution. It is typically expressed in moles per liter (mol L⁻¹) or moles per decimeter cubed (mol dm⁻³). A more concentrated solution contains a larger amount of solute per unit volume compared to a less concentrated solution.
Calculating Concentration:
Concentration (C) = moles of solute (n) / volume of solution (V)
To calculate the concentration of a solution, you need to know the number of moles of the solute and the volume of the solution in which the solute is dissolved. The unit of measurement for concentration is often mol dm⁻³ or mol L⁻¹.
Concentration and Titrations:
Titrations are a common laboratory technique used to determine the concentration of an unknown solution, typically an acid or a base, by reacting it with a solution of known concentration, called the titrant. In acid-base titrations, the reaction continues until the point of neutralization is reached, where the number of moles of the acid is stoichiometrically equivalent to the number of moles of the base.
The relationship between concentration and titrations can be demonstrated using the following formula:
moles of solute (n) = concentration (C) × volume (V)
This formula allows you to calculate the number of moles of solute in both the unknown solution and the titrant. By using stoichiometry, you can determine the concentration of the unknown solution.
For example, in an acid-base titration, the balanced chemical equation provides the stoichiometric ratio between the acid and the base. By knowing the concentration and volume of the titrant, you can calculate the moles of solute in the titrant. Using the stoichiometric ratio, you can then determine the moles of solute in the unknown solution. Finally, by dividing the moles of solute by the volume of the unknown solution, you can calculate its concentration.
Understanding concentration is crucial in chemistry, as it provides insights into the behavior of solutions and their reactions. Concentration plays a significant role in titrations, allowing chemists to determine the concentration of unknown solutions through stoichiometric relationships. By mastering the concept of concentration and its connection to titrations, you can tackle a wide range of chemical problems with confidence.
Simple Concentration Calculations
- If you dissolve 0.5 moles of sodium chloride (NaCl) in enough water to make a 1.0 dm³ solution, what is the concentration of the solution in mol dm⁻³?
- You have a 2.0 dm³ solution containing 1.0 moles of potassium nitrate (KNO₃). What is the concentration of the solution in mol dm⁻³?
- Calculate the concentration of a solution in mol dm⁻³, if you dissolve 0.75 moles of glucose (C₆H₁₂O₆) in water to make a 1.5 dm³ solution.
- You prepare a 0.5 dm³ solution containing 0.25 moles of sulfuric acid (H₂SO₄). What is the concentration of the solution in mol dm⁻³?
- A 3.0 dm³ solution contains 1.5 moles of acetic acid (CH₃COOH). Calculate the concentration of the solution in mol dm⁻³.
- Calculate the concentration of a solution in mol dm⁻³, if you dissolve 0.4 moles of calcium chloride (CaCl₂) in water to make a 2.0 dm³ solution.
- You have a 1.0 dm³ solution containing 2.0 moles of nitric acid (HNO₃). What is the concentration of the solution in mol dm⁻³?
- If you dissolve 0.6 moles of magnesium sulfate (MgSO₄) in enough water to make a 1.5 dm³ solution, what is the concentration of the solution in mol dm⁻³?
- Calculate the concentration of a solution in mol dm⁻³, if you dissolve 0.3 moles of ammonium chloride (NH₄Cl) in water to make a 0.5 dm³ solution.
- You prepare a 1.5 dm³ solution containing 0.75 moles of sodium hydroxide (NaOH). What is the concentration of the solution in mol dm⁻³?
Calculations that require unit conversions or further calculations
- If you dissolve 29.2 g of sodium chloride (NaCl) in enough water to make a 200 cm³ solution, what is the concentration of the solution in mol dm⁻³? (Molar mass of NaCl = 58.4 g/mol)
- You have a 500 cm³ solution containing 20.0 g of potassium nitrate (KNO₃). What is the concentration of the solution in mol dm⁻³? (Molar mass of KNO₃ = 101.1 g/mol)
- Calculate the concentration of a solution in mol dm⁻³, if you dissolve 54.0 g of glucose (C₆H₁₂O₆) in water to make a 300 cm³ solution. (Molar mass of C₆H₁₂O₆ = 180.0 g/mol)
- You prepare a 100 cm³ solution containing 9.8 g of sulfuric acid (H₂SO₄). What is the concentration of the solution in mol dm⁻³? (Molar mass of H₂SO₄ = 98.0 g/mol)
- A 150 cm³ solution contains 18.0 g of acetic acid (CH₃COOH). Calculate the concentration of the solution in mol dm⁻³. (Molar mass of CH₃COOH = 60.0 g/mol)
- Calculate the concentration of a solution in mol dm⁻³, if you dissolve 11.1 g of calcium chloride (CaCl₂) in water to make a 250 cm³ solution. (Molar mass of CaCl₂ = 110.9 g/mol)
- You have a 400 cm³ solution containing 12.8 g of nitric acid (HNO₃). What is the concentration of the solution in mol dm⁻³? (Molar mass of HNO₃ = 63.0 g/mol)
- If you dissolve 36.0 g of magnesium sulfate (MgSO₄) in enough water to make a 600 cm³ solution, what is the concentration of the solution in mol dm⁻³? (Molar mass of MgSO₄ = 120.4 g/mol)
- Calculate the concentration of a solution in mol dm⁻³, if you dissolve 10.7 g of ammonium chloride (NH₄Cl) in water to make a 200 cm³ solution. (Molar mass of NH₄Cl = 53.5 g/mol)
- You prepare a 500 cm³ solution containing 20.0 g of sodium hydroxide (NaOH). What is the concentration of the solution in mol dm⁻³? (Molar mass of NaOH = 40.0 g/mol)
Titrations
Example: Candy is performing a titration to determine the concentration of a hydrochloric acid (HCl) solution. She uses a 25.0 cm³ aliquot of the HCl solution and titrate it with a 0.100 mol dm⁻³ sodium hydroxide (NaOH) solution. The balanced chemical equation for the reaction is:
HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)
It takes 35.0 cm³ of the sodium hydroxide solution to completely neutralize the hydrochloric acid. Calculate the concentration of the hydrochloric acid solution in mol dm⁻³.
| Chemical Name | Mole Ratio | Moles | Volume (dm³) | Concentration (mol dm⁻³) |
|---|---|---|---|---|
| HCl | 1 | 0.025 | ? | |
| NaOH | 1 | 0.035 | 0.100 |
From this I can see that I can solve the number of moles of HCl = 0.035 x 0.100 = 0.0035
Because the ratio is 1:1 that must mean that the same number of moles of NaOH was used to neutralise the acid
| Chemical Name | Mole Ratio | Moles | Volume (dm³) | Concentration (mol dm⁻³) |
|---|---|---|---|---|
| HCl | 1 | 0.0035 | 0.025 | ? |
| NaOH | 1 | 0.0035 | 0.035 | 0.100 |
Now we have the number of moles and the volume so we can find the concentration of the NaOH
concentration = 0.0035 / 0.025
Concentration = 0.14 mol dm-3
21- Lynn is performing a titration to determine the concentration of an acetic acid (CH₃COOH) solution. She uses a 50.0 cm³ aliquot of the acetic acid solution and titrate it with a 0.150 mol dm⁻³ sodium hydroxide (NaOH) solution. The balanced chemical equation for the reaction is:
CH₃COOH(aq) + NaOH(aq) → CH₃COONa(aq) + H₂O(l)
It takes 45.0 cm³ of the sodium hydroxide solution to completely neutralize the acetic acid. Calculate the concentration of the acetic acid solution in mol dm⁻³.
22- Rico wants to determine the concentration of a sulfuric acid (H₂SO₄) solution. He takes a 20.0 cm³ aliquot of the sulfuric acid solution and titrate it with a 0.200 mol dm⁻³ potassium hydroxide (KOH) solution. The balanced chemical equation for the reaction is:
H₂SO₄(aq) + 2KOH(aq) → K₂SO₄(aq) + 2H₂O(l)
It takes 55.0 cm³ of the potassium hydroxide solution to completely neutralize the sulfuric acid. Calculate the concentration of the sulfuric acid solution in mol dm⁻³.
23- Song Song is performing a titration to determine the concentration of a phosphoric acid (H₃PO₄) solution. She uses a 10.0 cm³ aliquot of the phosphoric acid solution and titrate it with a 0.050 mol dm⁻³ calcium hydroxide (Ca(OH)₂) solution. The balanced chemical equation for the reaction is:
2H₃PO₄(aq) + 3Ca(OH)₂(aq) → Ca₃(PO₄)₂(s) + 6H₂O(l)
It takes 60.0 cm³ of the calcium hydroxide solution to completely neutralize the phosphoric acid. Calculate the concentration of the phosphoric acid solution in mol dm⁻³.
24- Freddy wants to determine the concentration of a nitric acid (HNO₃) solution. He takes a 30.0 cm³ aliquot of the nitric acid solution and titrate it with a 0.250 mol dm⁻³ barium hydroxide (Ba(OH)₂) solution. The balanced chemical equation for the reaction is:
2HNO₃(aq) + Ba(OH)₂(aq) → Ba(NO₃)₂(aq) + 2H₂O(l)
It takes 40.0 cm³ of the barium hydroxide solution to completely neutralize the nitric acid. Calculate the concentration of the nitric acid solution in mol dm⁻³.
Answers
- 0.5 mol dm⁻³ (0.5 moles in 1.0 dm³).
- 0.5 mol dm⁻³ (1.0 moles in 2.0 dm³).
- 0.5 mol dm⁻³ (0.75 moles in 1.5 dm³).
- 0.5 mol dm⁻³ (0.25 moles in 0.5 dm³).
- 0.5 mol dm⁻³ (1.5 moles in 3.0 dm³).
- 0.2 mol dm⁻³ (0.4 moles in 2.0 dm³).
- 2.0 mol dm⁻³ (2.0 moles in 1.0 dm³).
- 0.4 mol dm⁻³ (0.6 moles in 1.5 dm³).
- 0.6 mol dm⁻³ (0.3 moles in 0.5 dm³).
- 0.5 mol dm⁻³ (0.75 moles in 1.5 dm³).
- 2.5 mol dm⁻³ (0.5 moles of NaCl in 0.2 dm³).
- 0.4 mol dm⁻³ (0.197 moles of KNO₃ in 0.5 dm³).
- 1.0 mol dm⁻³ (0.3 moles of C₆H₁₂O₆ in 0.3 dm³).
- 1.0 mol dm⁻³ (0.1 moles of H₂SO₄ in 0.1 dm³).
- 2.0 mol dm⁻³ (0.3 moles of CH₃COOH in 0.15 dm³).
- 0.4 mol dm⁻³ (0.1 moles of CaCl₂ in 0.25 dm³).
- 0.508 mol dm⁻³ (0.203 moles of HNO₃ in 0.4 dm³).
- 0.5 mol dm⁻³ (0.3 moles of MgSO₄ in 0.6 dm³).
- 1.0 mol dm⁻³ (0.2 moles of NH₄Cl in 0.2 dm³).
- 1.0 mol dm⁻³ (0.5 moles of NaOH in 0.5 dm³).
- acetic acid solution is 0.135 mol dm⁻³.
- sulfuric acid solution is 0.275 mol dm⁻³.
- phosphoric acid solution is 0.2 mol dm⁻³.
- nitric acid solution is 0.67 mol dm⁻³.
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