The balanced equation of this reaction:
The total amount of moles of barium hydroxide is:
n(Ba(OH)2)=C(Ba(OH)2)∗V(Ba(OH)2)=0.1175M∗0.03146L=0.0037moles According to stoichoiometric quotients, there are 2 moles of nitric acid per each mole of barium hydroxide, so we get:
n(HNO3)=2n(Ba(OH)2)=0.0037∗2moles=0.0074moles According to molar concentration formula:
CM(HNO3)=Vn=0.0074moles/0.15L=0.0493M Answer: the concentration of the nitric acid in the sample is 0.0493M.
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