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What I reason: Calcium chloride is the salt of hydrochloric acid and calcium hydroxide. Calcium hydroxide is usually not considered a strong base, and I believe this is because of it's low solubility. $\ce{HCl}$ is a strong acid, and so the salt should be slightly acidic. Wikipedia states the $\mathrm{p}K_\mathrm{a}$ of $\ce{CaCl2}$ is between 8-9, which is in fact slightly acidic, confirming my theory.

What I've experienced: I've prepared a number of aqueous calcium chloride solutions from distilled water, all of which are purple in the presence of universal indicator and tested with a calibrated $\mathrm{pH}$ probe to be basic. I've tried multiple sources of calcium chloride and each is basic.

TLDR: Theoretically and literature values state that $\ce{CaCl2}$ is acidic, but my empirical evidence shows that it is basic.

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Calcium hydroxide has solubility about 1.9 g/L. This is enough to create pH above 11, i.e. a strongly basic solution.

$\ce{CaCl2}$ solutions should be very slightly acidic if they were made from pure $\ce{CaCl2}$. This might not be the case. In industry calcium chloride is produces by reaction of calcium hydroxide with ammonium chloride, so industrial-grade calcium chloride is likely to be contaminated with calcium hydroxide.

Another potential reason for contamination is hot drying. Calcium chloride is quite hygroscopic, so it needs to be dried before use. When heated to hight enough temperature, moist calcium chloride hydrolyzes https://link.springer.com/article/10.1007/BF02654424

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  • $\begingroup$ "Calcium hydroxide has solubility about 1.9 g/L. This is enough to create pH about 10-11" I appreciate to solve and explain it.thanks. $\endgroup$
    – Adnan AL-Amleh
    May 10 '19 at 17:30
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    $\begingroup$ @AdnanAL-Amleh pH+pOH is roughly 14. For Ca(OH)2 solution [OH] is roughly twice molar concentration of calcium hydroxide, which is roughly 0.025 for saturated solution. $pOH = -log [OH]$; $log_{10} 0.025 = 1.6$; $14-1.6 > 12$. You don't need much base to create a strongly basic solution. $\endgroup$
    – permeakra
    May 10 '19 at 20:48
  • $\begingroup$ You mean : $\ce{Ca(OH)2_\mathrm{(aq)} <=>Ca(OH)^+_\mathrm{(aq)} + OH^-_\mathrm{(aq)}}$ , so: $$[\ce{OH-}]=[\ce{Ca(OH)2}] = \frac{1.9}{74}=\pu{0.025M}$$ $\endgroup$
    – Adnan AL-Amleh
    May 10 '19 at 22:27
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    $\begingroup$ @AdnanAL-Amleh Yes. 0.025 M for $[OH-] $ is a low estimate, it is probably higher because $\ce{Ca(OH)+}$ can dissociate too. $\endgroup$
    – permeakra
    May 10 '19 at 22:55

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