By titration it is found that 69.9 mL of 0.151 M NaOH(aq) is needed to neutralize 25.0 mL of HCl(aq). Calculate the concentration of the HCl solution.

mL1 x M1 = mL2 x M2

This equation works because the mols HCl = mols NaOH.

Well, well, well, looks like we've got a chemistry question here! Don't worry, I'm here to help.

To find the concentration of the HCl solution, we first need to understand the concept of stoichiometry. Are you familiar with it, or should I give you a crash course?

To calculate the concentration of the HCl solution, you can use the equation:

M1V1 = M2V2

where:
- M1 is the concentration of the NaOH solution,
- V1 is the volume of the NaOH solution,
- M2 is the concentration of the HCl solution, and
- V2 is the volume of the HCl solution.

Given:
M1 = 0.151 M
V1 = 69.9 mL
V2 = 25.0 mL

Now, let's substitute these values into the equation:

(0.151 M)(69.9 mL) = M2(25.0 mL)

Solving for M2:

M2 = (0.151 M)(69.9 mL) / 25.0 mL

M2 = 0.4213492 M

Rounding to the appropriate number of significant figures, the concentration of the HCl solution is approximately 0.421 M.

To calculate the concentration of the HCl solution, we can use the formula:

M1V1 = M2V2

Where:
M1 = concentration of NaOH solution (in M)
V1 = volume of NaOH solution used (in mL)
M2 = concentration of HCl solution (in M)
V2 = volume of HCl solution used (in mL)

From the problem statement, we have:

M1 = 0.151 M
V1 = 69.9 mL
V2 = 25.0 mL

Plugging in the values into the formula, we get:

0.151 M x 69.9 mL = M2 x 25.0 mL

Now, we can solve for M2:

M2 = (0.151 M x 69.9 mL) / 25.0 mL

M2 = 0.419 M

Therefore, the concentration of the HCl solution is 0.419 M.