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  1. Jul 2020
  2. chem.libretexts.org chem.libretexts.org
    Neutralization
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    1. AAAthhzz 27 Jul 2020
      V2=M1V1M2=(0.0452)(0.045)0.0955=21.2mL(20)(20)V2=M1V1M2=(0.0452)(0.045)0.0955=21.2mLV_2= \dfrac{M_1V_1}{M_2} = \dfrac{(0.0452)(0.045)}{0.0955} = 21.2\; mL Therefore it takes 21.2 mL of Ba(OH)2Ba(OH)2Ba(OH)_2 to titrate 45.00 mL HNO3HNO3HNO_3.

      THIS PART IS wRONG, Ba(OH)2 has two OH- groups so needs to be divided by 2

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    chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Acids_and_Bases/Acid//Base_Reactions/Neutralization