What volume (mL)of 0.102 M NaOH is required to neutralize 17.1 mL of 0.443 M HCl?
A)74.3
B)3.94
C)0.773
D)0.0135
E)0.000773

Name: What volume (mL)of 0.102 M NaOH is required to neutralize 17.1 mL of 0.443 M HCl? A)74.3 B)3.94 C)0.773 D)0.0135 E)0.000773
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Quizplus · Accepted Answer

To neutralize the acid, use the formula M1V1 = M2V2. Here, M1 = 0.443 M, V1 = 17.1 mL, M2 = 0.102 M. Solving for V2 gives V2 = (0.443 * 17.1) / 0.102 = 74.3 mL.

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