How many milliliters of commercial hydrochloric acid, which is 11.6 M, should be used to prepare 6.94 L of 0.52 M HCl?
To solve this problem, we can use the equation:
\( M_1V_1 = M_2V_2 \)
where:
- \( M_1 \) is the initial concentration of the acid (11.6 M)
- \( V_1 \) is the volume of the initial solution (unknown)
- \( M_2 \) is the final concentration of the acid (0.52 M)
- \( V_2 \) is the final desired volume of the solution (6.94 L)
To find the volume of the initial solution (\( V_1 \)), we rearrange the equation:
\( V_1 = \frac{{M_2V_2}}{{M_1}} \)
Now, we can substitute the given values into the equation:
\( V_1 = \frac{{0.52 \, \text{M}} \times 6.94 \, \text{L}}}{{11.6 \, \text{M}}} \)
Calculating this:
\( V_1 = 0.52 \times \frac{{6.94}}{{11.6}} \, \text{L} \)
\( V_1 = 0.31 \, \text{L} \)
To convert liters to milliliters, we multiply by 1000:
\( V_1 = 0.31 \, \text{L} \times 1000 \, \text{mL/L} \)
\( V_1 = 310 \, \text{mL} \)
Therefore, you will need 310 milliliters of the 11.6 M hydrochloric acid to prepare 6.94 L of 0.52 M HCl.