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Answer to Question #37714 in Inorganic Chemistry for Emily

Question #37714
0.965g of aluminium nitrate is dissolved in 125 mL of water. What is the molarity of the cation and anion in the resulting solution?
Expert's answer
Al(NO3)3 -> Al3+ + 3(NO3)-

m(Al(NO3)3) = 0.965g
V(H2O ) = 125 ml = 0.125L
M(Al(NO3)3) = 212.996
n(Al(NO3)3) = m/M = 0.965g / 213.0 = 0.004530 mol
n(Al3+) = n(Al(NO3)3) = 0.00453 mol
n[(NO3)-] = 3*n(Al(NO3)3) = 0.0136 mol
c(Al3+) = n/V = 0.0362 mol/l
n[(NO3)-] = n/V = 0.109 mol/l

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