Question

A 20.0mL sample of 0.850 MHCl is titrated with NaOH.If 18.2mL of NaOH is required to reach the equivalence point, what is the molarity of the NaOH?

A) 0.932MNaOH
B) 0.932mol/L
C) 1.70MNaOH
D) 1.70mol/L

Answer 1

To find the molarity of NaOH in a titration with HCl, we use the formula M1V1 = M2V2, which gives us the molarity of NaOH as 0.932 M, therefore the correct answer is A) 0.932 M NaOH.

Step-by-step explanation:

The student's question pertains to finding the molarity of a NaOH solution that has been titrated against a known concentration and volume of HCl solution. To find the molarity of NaOH, we use the titration formula:

M1 V1 = M2 V2

Where M1 and V1 are the molarity and volume of HCl, and M2 and V2 are the molarity and volume of NaOH.

Using the given volumes and the molarity of HCl (0.850 M for 20.0 mL) and NaOH (18.2 mL) required to reach the equivalence point, we can solve for the molarity of NaOH (M2):

0.850 M × 20.0 mL = M2 × 18.2 mL

M2 = (0.850 M × 20.0 mL) / (18.2 mL)

M2 = 0.932 M NaOH

Thus, the correct answer is A) 0.932 M NaOH.