A student prepares 150.0 mL of 1.40 M HCl using 35.0 mL of a stock solution. What is the concentration of the stock solution?

Use [tex]\( M_i V_i = M_f V_f \)[/tex].

A. 0.327 M
B. 4.29 M
C. 6.00 M
D. 7.35 M



Answer :

To determine the concentration [tex]\( M_j \)[/tex] of the stock solution, we can use the dilution formula given by:

[tex]\[ M_j \times V_j = M_f \times V_f \][/tex]

Where:
- [tex]\( M_j \)[/tex] is the initial molarity (concentration) of the stock solution.
- [tex]\( V_j \)[/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:
- Final volume [tex]\( V_f = 150.0 \)[/tex] mL
- Final molarity [tex]\( M_f = 1.40 \)[/tex] M
- Initial volume [tex]\( V_j = 35.0 \)[/tex] mL

We need to find the initial molarity [tex]\( M_j \)[/tex].

Using the formula, we have:

[tex]\[ M_j \times 35.0 \text{ mL} = 1.40 \text{ M} \times 150.0 \text{ mL} \][/tex]

Rearranging to solve for [tex]\( M_j \)[/tex], we get:

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

Calculating the right side:

[tex]\[ M_j = \frac{210.0 \text{ M} \cdot \text{mL}}{35.0 \text{ mL}} \][/tex]
[tex]\[ M_j = 6.00 \text{ M} \][/tex]

Hence, the concentration of the stock solution is [tex]\( 6.00 \)[/tex] M.

Therefore, the answer is:
[tex]\[ 6.00 \text{ M} \][/tex]