We’re being asked to **calculate the molarity of NaOH (strong base)** from the **given the molarity and volume of H _{2}SO_{4}**

This means that **25.00 mL of NaOH **required** ****39.1 mL of 0.112 M H _{2}SO_{4}** to reach the equivalence point.

What is the molarity of a NaOH solution if 39.1 mL of a 0.112 M H_{2}SO_{4} solution is required to neutralize a 25.0-mL sample of the NaOH solution (as in a titration)?

A) 54.7 M

B) 0.224 M

C) 0.350 M

D) 0.143 M

E) 0.0716 M

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