A 25.00 mL aliquot of HCl of unknown concentration is titrated with 25.62 mL of 0.1011 M NaOH. What is the molarity of the HCl solution?

To find the molarity of the HCl solution, we can use the concept of stoichiometry and perform a simple calculation.

First, we need to determine the number of moles of NaOH used in the titration:

Moles of NaOH = Volume of NaOH (in L) × Molarity of NaOH

Converting the volume of NaOH from milliliters (mL) to liters (L):

Volume of NaOH = 25.62 mL ÷ 1000 mL/L = 0.02562 L

Now we can calculate the moles of NaOH:

Moles of NaOH = 0.02562 L × 0.1011 mol/L = 0.002598 mol

Since the balanced chemical equation between NaOH and HCl is 1:1, the moles of NaOH used will be equal to the moles of HCl present in the aliquot.

Thus, the moles of HCl = moles of NaOH = 0.002598 mol

Next, we need to determine the molarity of the HCl solution:

Molarity of HCl = Moles of HCl ÷ Volume of HCl (in L)

Converting the volume of HCl from milliliters (mL) to liters (L):

Volume of HCl = 25.00 mL ÷ 1000 mL/L = 0.02500 L

Now we can calculate the molarity of the HCl solution:

Molarity of HCl = 0.002598 mol ÷ 0.02500 L ≈ 0.1039 M

Therefore, the molarity of the HCl solution is approximately 0.1039 M.

To determine the molarity (M) of the HCl solution, you can use the equation:

M1 x V1 = M2 x V2

Where:
M1 is the molarity of the NaOH solution
V1 is the volume of the NaOH solution used in the titration
M2 is the molarity of the HCl solution
V2 is the volume of the HCl solution

Given:
M1 = 0.1011 M (molarity of NaOH solution)
V1 = 25.62 mL (volume of NaOH solution used)
V2 = 25.00 mL (volume of HCl solution)

Plug in the values into the equation:

0.1011 M x 25.62 mL = M2 x 25.00 mL

Solve for M2:

M2 = (0.1011 M x 25.62 mL) / 25.00 mL

M2 = 0.1033 M

Therefore, the molarity of the HCl solution is approximately 0.1033 M.