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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

Respuesta :

helona

Answer:

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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