To determine the concentration of a solution after dilution, we use the dilution equation:
[tex]\[ M_1 \times V_1 = M_2 \times V_2 \][/tex]
where:
- [tex]\( M_1 \)[/tex] is the initial molarity of the solution
- [tex]\( V_1 \)[/tex] is the initial volume of the solution
- [tex]\( M_2 \)[/tex] is the final molarity of the solution after dilution
- [tex]\( V_2 \)[/tex] is the final volume of the solution after dilution
Given the problem:
- [tex]\( M_1 = 2.03 \, \text{M} \)[/tex]
- [tex]\( V_1 = 93 \, \text{mL} \)[/tex]
- [tex]\( V_2 = V_1 + \text{volume of water added} = 93 \, \text{mL} + 3920 \, \text{mL} = 4013 \, \text{mL} \)[/tex]
We need to find [tex]\( M_2 \)[/tex].
Step-by-step solution:
1. Write down the dilution equation:
[tex]\[
M_1 \times V_1 = M_2 \times V_2
\][/tex]
2. Substitute the known values into the equation:
[tex]\[
2.03 \, \text{M} \times 93 \, \text{mL} = M_2 \times 4013 \, \text{mL}
\][/tex]
3. To isolate [tex]\( M_2 \)[/tex], divide both sides of the equation by [tex]\( V_2 \)[/tex]:
[tex]\[
M_2 = \frac{2.03 \times 93}{4013}
\][/tex]
4. Calculate the right-hand side:
[tex]\[
M_2 = \frac{188.79}{4013}
\][/tex]
5. Perform the division:
[tex]\[
M_2 \approx 0.047
\][/tex]
6. Therefore, the molarity of the diluted solution is approximately:
[tex]\[
M_2 \approx 0.047 \, \text{M}
\][/tex]
So the molarity of the diluted KF solution is approximately [tex]\( 0.047 \, \text{M} \)[/tex].
