A student prepares [tex]150.0 \, \text{mL}[/tex] of [tex]1.40 \, M \, \text{HCl}[/tex] using [tex]35.0 \, \text{mL}[/tex] of a stock solution. What is the concentration of the stock solution?

Use [tex]M_i V_i = M_f V_f[/tex].

A. [tex]0.327 \, M[/tex]
B. [tex]4.29 \, M[/tex]
C. [tex]6.00 \, M[/tex]
D. [tex]7.35 \, M[/tex]



Answer :

To find the concentration of the stock solution (denoted as [tex]\( M_i \)[/tex]), we can use the dilution equation:

[tex]\[ M_i \times V_i = M_f \times V_f \][/tex]

where:
- [tex]\( M_i \)[/tex] is the initial molarity (concentration) of the stock solution,
- [tex]\( V_i \)[/tex] is the initial volume of the stock solution,
- [tex]\( M_f \)[/tex] is the final molarity (concentration) of the diluted solution,
- [tex]\( V_f \)[/tex] is the final volume of the diluted solution.

Given data:
- [tex]\( M_f = 1.40 \)[/tex] M (final molarity),
- [tex]\( V_f = 150.0 \)[/tex] mL (final volume),
- [tex]\( V_i = 35.0 \)[/tex] mL (initial volume).

We need to find [tex]\( M_i \)[/tex]. Rearrange the dilution equation to solve for [tex]\( M_i \)[/tex]:

[tex]\[ M_i = \frac{M_f \times V_f}{V_i} \][/tex]

Now, substitute the given values into the equation:

[tex]\[ M_i = \frac{1.40 \, \text{M} \times 150.0 \, \text{mL}}{35.0 \, \text{mL}} \][/tex]

Perform the multiplication in the numerator:

[tex]\[ 1.40 \, \text{M} \times 150.0 \, \text{mL} = 210.0 \, \text{M} \cdot \text{mL} \][/tex]

Then, divide by the initial volume:

[tex]\[ M_i = \frac{210.0 \, \text{M} \cdot \text{mL}}{35.0 \, \text{mL}} \][/tex]

[tex]\[ M_i = 6.0 \, \text{M} \][/tex]

So, the concentration of the stock solution is:

[tex]\[ 6.00 \, \text{M} \][/tex]

Therefore, the correct answer is [tex]\( 6.00 \, \text{M} \)[/tex].