Certainly! To answer your question thoroughly, I'll break it into parts and explain each concept before addressing the main calculation.
### Definitions:
1. Decinormal Solution:
- Normality (N): Normality is a measure of concentration equivalent to molarity (M) but takes into account the reactive capacity of the substance in reactions. It is defined as the number of equivalents of solute per liter of solution.
- Decinormal Solution (0.1 N): A decinormal solution is one-tenth of the normality unit, i.e., it contains 0.1 equivalents per liter of the solution.
2. Indicator:
- An indicator is a substance used in titrations to signal the end point of a reaction by changing color. It helps in determining when the reactants have been mixed in the proper proportions.
3. End Point:
- The end point is the stage in a titration at which the indicator changes color, signaling that the reaction is complete or has reached a significant stage. It closely approximates the stoichiometric equivalence point of the reaction.
### Calculation of Sodium Carbonate Required:
To prepare a decinormal solution of sodium carbonate ([tex]\(\text{Na}_2\text{CO}_3\)[/tex]) in 250 mL of solution, follow these steps:
1. Molar Mass of Sodium Carbonate ([tex]\(\text{Na}_2\text{CO}_3\)[/tex]):
- Sodium ([tex]\(\text{Na}\)[/tex]) has an atomic mass of ~22.99 g/mol.
- Carbon ([tex]\(\text{C}\)[/tex]) has an atomic mass of 12.01 g/mol.
- Oxygen ([tex]\(\text{O}\)[/tex]) has an atomic mass of 16.00 g/mol.
- The molar mass of sodium carbonate is thus calculated as:
[tex]\[
\text{Molar Mass of Na}_2\text{CO}_3 = (2 \times 22.99) + 12.01 + (3 \times 16.00) = 105.99 \, \text{g/mol}
\][/tex]
2. Equivalent Weight of Sodium Carbonate:
- Sodium carbonate [tex]\(\text{Na}_2\text{CO}_3\)[/tex] dissociates into 2 [tex]\(\text{Na}^+\)[/tex] and 1 [tex]\(\text{CO}_3^{2-}\)[/tex].
- In this context, the equivalent weight of sodium carbonate is half its molar mass:
[tex]\[
\text{Equivalent Weight of Na}_2\text{CO}_3 = \frac{\text{Molar Mass of Na}_2\text{CO}_3}{2} = \frac{105.99}{2} = 52.995 \, \text{g/equiv}
\][/tex]
3. Calculating the Mass of Sodium Carbonate Needed:
- To prepare a decinormal solution ([tex]\(0.1 \, N\)[/tex]) in 250 mL ([tex]\(0.25 \, L\)[/tex]), the formula for mass required is:
[tex]\[
\text{Mass} = \text{Normality} \times \text{Equivalent Weight} \times \text{Volume}
\][/tex]
- Substituting in the known values:
[tex]\[
\text{Mass} = 0.1 \, \text{N} \times 52.995 \, \text{g/equiv} \times 0.25 \, \text{L} = 1.324875 \, \text{g}
\][/tex]
### Conclusion:
To prepare a decinormal (0.1 N) solution of sodium carbonate in 250 mL of water, you need precisely 1.324875 grams of sodium carbonate ([tex]\(\text{Na}_2\text{CO}_3\)[/tex]).
