What volume (mL)of 0.135 M NaOH is required to neutralize 13.7 mL of 0.129 M HCl?
A)13.1
B)0.24
C)14.3
D)0.076
E)6.55

Name: What volume (mL)of 0.135 M NaOH is required to neutralize 13.7 mL of 0.129 M HCl? A)13.1 B)0.24 C)14.3 D)0.076 E)6.55
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Quizplus · Accepted Answer

To neutralize the acid, the moles of NaOH must equal the moles of HCl. Moles of HCl = 0.129 M * 13.7 mL = 1.7673 mmol. The volume of NaOH needed = 1.7673 mmol / 0.135 M = 13.1 mL.

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