I was in the chemistry lab earlier this week. We were doing a qualitative experiment (no concentration given for any solution).

I added an unknown solution that contains $\ce{Ba^2+},$ $\ce{Ca^2+},$ and $\ce{Mg^2+}$ with $\ce{(NH4)2SO4},$ I got barium precipitate and added $\ce{(NH4)2C2O4}$ and another precipitate appeared. I was told the second precipitate is $\ce{Ca^2+}.$

But I recall from solubility mnemonic rule, I should be able to see precipitate from both $\ce{Ca^2+}$ and $\ce{Ba^2+}$ after adding ammonium sulfate. Why didn't I? What are the mechanics behind this? How did calcium just sit there as a spectator ion?

  • $\begingroup$ Can you please tell, where did you get this handout from, Can you please share other helpful ones if possible. Please !! $\endgroup$ Oct 11, 2019 at 20:05
  • $\begingroup$ Do you mean the CASH N' Gia Solubility Rule sheet link? I saw the mnemonic at reddit a while back and googled it. This is what I found. I think there's another sheet for precipitation rule sheet called "CCOPS", which is developed by UTexas at El Paso. $\endgroup$
    – Molly_K
    Oct 12, 2019 at 2:19
  • $\begingroup$ Here it is: CCOPS are insoluble salts: C-Carbonates and Chromates, O-OH (hydroxides), P-Phosphates, and S-Sulfides. In using the full mnemonic, PLs introduce sets of exceptions as HAPpy: HAP for Hg, Ag, Pb as exceptions for CASH N’ Gia solubility for halides; and as HAPpy (Hg, Ag, Pb) and CBS (Ca, Ba, Sr) for sulfates. see here: researchgate.net/publication/… $\endgroup$
    – Molly_K
    Oct 12, 2019 at 2:20

1 Answer 1


While calcium sulfate is usually termed insoluble, it is not the ‘sitting at the bottom like a rock’ type insoluble; rather, it is the ‘there’s no practical way for me to get the two ions into the same solution without precipitation, but I’m still able to identify both ions in solution’ type insoluble. Those sentences don’t really help, so let’s look at numbers.

\begin{array}{lcr} \hline \text{salt} & \text{mass solubility} & \text{molar solubility} \\ \hline \ce{BaSO4} & \pu{2.45e-3 g/l} & \pu{1.05e-5 mol/l}\\ \ce{CaSO4} & \pu{2.1 g/l} & \pu{1.54e-2 mol/l}\\ \ce{CaC2O4} & \pu{6.7e-4 g/l} & \pu{5.23e-6 mol/l}\\ \hline \end{array}

To calculate your actual experiment, you would have to transform these values into $K_\text{sp}$ values and then calculate how much would precipitate at each given step. Without wanting to do that tedious calculation, you can still arrive at the general conclusion:

  • addition of sulphate precipitates practically all barium ions

  • addition of sulphate precipitates a large part of calcium ions

    but: a non-neglegible amount of $\ce{CaSO4}$ remains dissolved

  • addition of oxalate precipitates practically all the remaining calcium.

In fact, the solubilities of $\ce{BaSO4, SrSO4}$ and $\ce{CaSO4}$ are such that a saturated solution of $\ce{SrSO4}$ will always give precipitate when barium is added but never with calcium or strontium; while a saturated solution of $\ce{CaSO4}$ will give a precipitate if either strontium or barium is added but never with calcium. This can be used to identify the heaviest alkaline earth metal in an unknown sample.

  • $\begingroup$ Where can I find molar solubility chart? Thank you. $\endgroup$
    – Molly_K
    Oct 12, 2019 at 3:26
  • $\begingroup$ @Molly_K I found solubilities on Wikipedia. You can also Google, check your favourite textbook or ask if your library knows something. $\endgroup$
    – Jan
    Oct 12, 2019 at 8:16
  • 1
    $\begingroup$ Thank you for your answer. With the Ksp and solubility table, all make sense now. I was quite thrown off when there's no concentration was given for any of the solutions. Ha. $\endgroup$
    – Molly_K
    Oct 13, 2019 at 3:47

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