I'm trying to determine the Solubility Constant $K_\mathrm{sp}$ of $\ce{Ba(NO3)2}$ dissolved in nitric acid solution $\ce{HNO3}$ from experimental data.

$\ce{HNO3}$ will be mostly dissociated so there are many $\ce{NO3-}$ ions already in the solvent before solid $\ce{Ba(NO3)2}$ is added.

When I calculate $K_\mathrm{sp} = [\ce{Ba^2+}][\ce{NO3-}]^2$, do I include in $[\ce{NO3-}]$ the ions already present in the solution before solid $\ce{Ba(NO3)2}$ is added?

For example, say I have dissolved ${0.1 mol}$ of barium nitrate in ${1 L}$ of ${1 M}$ nitric acid solution. There are now ${0.2 mol} + {1 mol} = {1.2 mol}$ of nitrate ion in the solution. Should my $K_\mathrm{sp}$ calculation be

$$K_\mathrm{sp} = 0.1\cdot 0.2^2,$$


$$K_\mathrm{sp} = 0.1\cdot 1.2^2?$$

I understand that the common ion effect in this case inhibits solubility somewhat, but it I think my professor told me to include ions from $\ce{HNO3}$ in the calculation, which seems wrong.


1 Answer 1


Yes, absolutely, all nitrate counts in solubility product regardless of source. The system cannot distinguish what a nitrate ion (or any dissociated species) was bonded to before it dissociated.

  • $\begingroup$ I get it now, thank you. My confusion was stemming from my misconception that the Ksp in HNO3 solution should be different from the Ksp in pure water. $\endgroup$
    – omzrs
    Commented Mar 29, 2019 at 2:52
  • 2
    $\begingroup$ @omzrs - At the risk of more adding confusion, you're not totally wrong. The Ksp actually depends on the chemical activity of the ions rather than their concentrations, and the activity is dependent on the ionic strength of the solution. Activity and ionic strength are concepts that you should study later. // I'll add that for a salt as soluble as barium nitrate the Ksp using concentration is really pretty useless. $\endgroup$
    – MaxW
    Commented Mar 29, 2019 at 3:27

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