How many moles of solid Ba(NO3)2 should be added to 300mL of 0.2M Fe(NO3)3 to increase the concentration of the NO3- ion to 1M? (assuming that the vol of the soltn remains the same)
How should I approach this problem? Thanks.
Answer:
Moles Fe(NO3)3 = 300 x 0.2 /1000 = 0.06
Moles NO3- = 0.06 x 3 = 0.18
M = moles / L
moles = M x L = 1 x 0.300 = 0.300
we have 0.18 moles NO3- so we add 0.300 - 0.18 = 0.12 mole NO3-
Ba(NO3)2 >> Ba2+ + 2NO3-
we need 0.12 /2 = 0.06 mole Ba(NO3)2
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