In a 1.0× 10–2 M solution of NH4ClO4(aq), identify the relative molar amounts of these species.
below is correct
The compound will completely ionize so that [ClO4-] is 1.0X10^-2 M. Because this is the conjugate base of a strong acid, [HClO4] = 0 M
Now, NH4+ is a weak acid that ionizes by the equation:
NH4+<--> NH3 + H+
Ka = [H+][NH3]/[NH4+] = 5.6X10^-10
Let [H+] = [NH3] = x, and [NH4+] = 0.010 – x. Because Ka is so small, x will be negligible compared to 0.010. So:
Ka = 5.6X10^-10 = x^2 / 0.010
x = [H+] = [NH3] = 2.37X10^-6 M, and [NH4+] = 0.010 M
[OH-] = 1.0X10^-14 / [H+] = 4.23X10^-9 M
Since water is the solvent, and since this is a fairly dilute solution, [H2O] = 55.6 M (1000 g/L / 18.0 g/mol) = 55.6 M H2O
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