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5 November, 20:11

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

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  1. 5 November, 22:31
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    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
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