A 50.0 ml sample of 0.0686 M AgNO3(aq) is added to 50.0 ml of 0.100 M NalO3(aq).

Question

A 50.0 ml sample of 0.0686 M AgNO3(aq) is added to 50.0 ml of 0.100 M NalO3(aq).
Calculate the [Ag+] at equilbrium in the resulting solution.
The Ksp value for AglO3(s) is 3.17x 10^-8.


[Ag+] ______ mol/L

Explanation / Answer

AgNO3 + NaIO3 -----> AgIO3 + NaNO3 no. of moles of AgNO3 = molarity X volume in litres = 0.0686 X 0.05 = 0.0034 no.of moles of NaIO3 = 0.05 X 0.1 = 0.005 as reaction is taking place in 1 : 1 ratio ....and NaIO3 is present in greater quantity so some of it will be left out ... amount of NaIO3 left out = 0.005 - 0.0034 = 0.0016 now total volume = 50 + 50 = 100 ml = 0.1 L conc. of NaIO3 left out = 0.0016/0.1 = 0.016 moles/L and as NaIO3 is a strong salt so it will be completely dissociated ... NaIO3 -----> Na+ + IO3- so [NaIO3] = [IO3-] = 0.016 moles/L AgIO3 Ag+ + IO3- Ksp = [Ag+] [IO3-] 3.17 X 10^-8 = [Ag+] X 0.016 [Ag+] = 3.17 X 10^-8/0.016 = 1.981 X 10^-6 moles/L feel free to ask any question ...

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