calculate the volume needed to make 2.20 x 10^2 ml of 0.100M cacl2. what would be the total concentration of the final (Cl-) alone?
To calculate the volume needed to make a specific concentration of a solution, we can use the formula:
\( C_1V_1 = C_2V_2 \)
Where:
C1 = Initial concentration
V1 = Initial volume
C2 = Final concentration
V2 = Final volume
In this case, we want to make 2.20 x 10^2 ml of a CaCl2 solution with a concentration of 0.100 M. Let's use the formula to calculate the volume needed.
Let's rearrange the formula to solve for V2:
\( V_2 = \frac{{C_1V_1}}{{C_2}} \)
Substituting the values given:
C1 = 0.100 M
V1 = Volume (unknown)
C2 = 2.20 x 10^2 ml
\( V_2 = \frac{{0.100 \, \text{M} \cdot V_1}}{{2.20 \cdot 10^2 \, \text{ml}}} \)
To calculate the volume needed, we need to rearrange the equation and solve for V1. Multiply both sides of the equation by \(\frac{{2.20 \cdot 10^2 \, \text{ml}}}{{0.1 \, \text{M}}} \):
\( V_1 = \frac{{V_2 \cdot C_2}}{{C_1}} \)
\( V_1 = \frac{{2.20 \cdot 10^2 \, \text{ml} \cdot 0.100 \, \text{M}}}{{0.100 \, \text{M}}} \)
\( V_1 = 2.20 \cdot 10^2 \, \text{ml} \)
Therefore, the volume needed to make 2.20 x 10^2 ml of 0.100 M CaCl2 is also 2.20 x 10^2 ml.
Now, to determine the total concentration of the final Cl- ions alone, we need to consider how many moles of Cl- are present in the solution. Since CaCl2 dissociates into three ions (1 Ca2+ and 2 Cl-), the concentration of Cl- alone is the double the concentration of CaCl2. Therefore, the final concentration of Cl- alone is:
Final concentration of Cl- = 2 x 0.100 M = 0.200 M
Hence, the total concentration of Cl- ions alone would be 0.200 M.