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

A) 0.932 M NaOH
B) 0.932 mol/L
C) 1.70 M NaOH
D) 1.70 mol/L

Answer :

Final answer:

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.

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.

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Rewritten by : Jeany