Ask Question
31 March, 15:24

25.0cm3 of arsenic acid, H3AsO4, required 37.5cm3 of 0.100 mol/dm3 sodium hydroxide, NaOH, for neutralization

3NaOH (aq) + H3AsO4 (aq) - > Na3AsO4 (aq) + 3 H2O (1)

Find the concentration of the acid in g/dm3

+4
Answers (1)
  1. 31 March, 18:22
    0
    H3ASO4 + 3NaOH - > Na3ASO4 + 3H2O

    1:3 reaction

    moles = vol/1000*M = 35.7/1000*0.100=0.00357moles of NaOH

    moles of H3ASO4 = 0.00357/3=0.00119

    concentration=moles/volume = 0.00119/25.0

    and then I'm not sure
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “25.0cm3 of arsenic acid, H3AsO4, required 37.5cm3 of 0.100 mol/dm3 sodium hydroxide, NaOH, for neutralization 3NaOH (aq) + H3AsO4 (aq) - > ...” in 📙 Chemistry if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers