Question

Concentration of solutions in molarity, dilution and titration (simple and back titration) with it's references

Answer 1

### Concentration of Solutions in Molarity

**Molarity (M)** is a measure of concentration expressed as the number of moles of solute per liter of solution. It can be calculated using the formula:

\[
\text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}}
\]

Where:
- Moles of solute = mass of solute (g) / molar mass of solute (g/mol).
- Liters of solution = volume of the solution in liters.

### Dilution

Dilution is the process of reducing the concentration of a solute in a solution, usually by adding more solvent. The relationship between the concentrations and volumes before and after dilution can be expressed using the dilution equation:

\[
C_1V_1 = C_2V_2
\]

Where:
- \(C_1\) is the initial concentration (before dilution),
- \(V_1\) is the initial volume,
- \(C_2\) is the final concentration (after dilution),
- \(V_2\) is the final volume.

### Titration

Titration is a quantitative analytical method used to determine the concentration of a solute in a solution. This is accomplished by adding a titrant (a solution of known concentration) to the analyte (the solution of unknown concentration) until the reaction reaches its endpoint, which is often indicated by a color change.

#### Simple Titration

In a simple titration, the volume of titrant required to reach the endpoint is measured, and using the stoichiometry of the reaction, the concentration of the analyte can be calculated. The formula used is:

\[
C_1V_1 = C_2V_2
\]

Where:
- \(C_1\) is the concentration of titrant,
- \(V_1\) is the volume of titrant used,
- \(C_2\) is the concentration of analyte,
- \(V_2\) is the volume of analyte.

**Example:**
If 25.0 mL of NaOH (0.1 M) is used to titrate 50.0 mL of HCl, what is the concentration of HCl?

1. Write the balanced reaction:

\( \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \)

This tells us the molar ratio is 1:1.

2. Calculate moles of NaOH used:
Moles of NaOH = 0.1 M × 0.025 L = 0.0025 moles
Since the stoichiometry is 1:1, moles of HCl = 0.0025 moles.

3. Calculate the concentration of HCl:
C = moles/volume = 0.0025 moles / 0.050 L = 0.050 M

#### Back Titration

Back titration is a technique used when the endpoint of a direct titration is difficult to identify or when the analyte or titrant is insoluble. In a back titration, an excess of a standard solution is added to the analyte and the unreacted portion of the standard solution is titrated with another solution.

**Steps:**
1. Add a known excess of titrant to the analyte.
2. Allow the reaction to go to completion.
3. Titrate the unreacted titrant with a second titrant.

**Example:**
Suppose you have an unknown amount of a carbonate, and you add 0.1 M HCl.
- You add 50 mL of HCl, which means you initially have:
Moles of HCl = 0.1 M × 0.050 L = 0.005 moles

- If you find that 30 mL of NaOH is needed to neutralize the excess HCl, calculate the moles of NaOH used. Applying \(C_1V_1 = C_2V_2\):

If NaOH is also 0.1 M, then moles of NaOH = 0.1 M × 0.030 L = 0.003 moles.

- The moles of HCl that reacted with the carbonate are:
\(0.005 - 0.003 = 0.002 \) moles.

- Therefore, you can determine the amount of the carbonate by knowing its molar equivalent.

### References
- "Chemistry: The Central Science" by Theodore L. Brown, H. Eugene LeMay, Bruce Eckardt, and Catherine Murphy.
- "Quantitative Chemical Analysis" by Daniel C. Harris.
- "General Chemistry" by Linus Pauling.
- Online resources: Websites like Khan Academy or educational chemistry resources also provide further explanations and illustrative examples.