11.83 mL of 0.5008 M NaOH was required to neutralize 25.00 mL of HCl with unknown...
11.83 mL of 0.5008 M NaOH was required to neutralize 25.00 mL of
HCl with unknown molarity.
How many moles per liter of NaOH was used?
Homework Answers
Moles per litre or Molarity
Which is given in the question = 0.5008 M
I calculate the molarity of HCl
M1V1 = M2V2
0.5008 * 11.83 = 25.00 * M2
M2 = 0.237
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