A 25.0 mL sample of an HCl solution is titrated with a 0.139 M NaOH solution. The equivalence point is reached with 15.4 mL of base. The concentration of HCl is ________ M.
A)11.7
B)0.00214
C)0.0856
D)0.267
E)0.139

Question

Quizplus · Accepted Answer

We can use the formula:

M1V1 = M2V2

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

At the equivalence point, the moles of HCl = moles of NaOH.

Therefore:

M1V1 = M2V2
M1(25.0 mL) = (0.139 M)(15.4 mL)

Solving for M1:

M1 = (0.139 M)(15.4 mL) / (25.0 mL)
M1 = 0.0856 M

Therefore, the concentration of HCl is 0.0856 M, which is choice C.

A 250 ML Sample of an HCl Solution Is Titrated with with a 0.139