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Is calcium chloride an acidic or basic salt?
Anthocyanins as pH-indicators: stop the titration at red, purple or blue?Hydrolysis of ammonium clorideWhy is CaCl2 called calcium chloride?finding cation of unknown salt in solution from precipitateWhy is NaHCO3 a basic salt?Is ciprofloxacin acidic or basic?What do we call ions without basic and acidic propertiesHow can I convert weight percentage of calcium chloride hexahydrate to calcium chloride dihydrate?Calcium-containing acidic solutionAcidic, Basic, and Neutral Medium
$begingroup$
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. $ceHCl$ is a strong acid, and so the salt should be slightly acidic. Wikipedia states the $mathrmpK_mathrma$ of $ceCaCl2$ 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 $mathrmpH$ probe to be basic. I've tried multiple sources of calcium chloride and each is basic.
TLDR:
Theoretically and literature values state that $ceCaCl2$ is acidic, but my empirical evidence shows that it is basic.
acid-base solutions ionic-compounds
edited May 11 at 6:26
Mathew Mahindaratne
7,7641030
asked May 10 at 12:53
124c41124c41
947
$endgroup$
add a comment |
$begingroup$
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. $ceHCl$ is a strong acid, and so the salt should be slightly acidic. Wikipedia states the $mathrmpK_mathrma$ of $ceCaCl2$ 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 $mathrmpH$ probe to be basic. I've tried multiple sources of calcium chloride and each is basic.
TLDR:
Theoretically and literature values state that $ceCaCl2$ is acidic, but my empirical evidence shows that it is basic.
acid-base solutions ionic-compounds
edited May 11 at 6:26
Mathew Mahindaratne
7,7641030
asked May 10 at 12:53
124c41124c41
947
$endgroup$
add a comment |
$begingroup$
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. $ceHCl$ is a strong acid, and so the salt should be slightly acidic. Wikipedia states the $mathrmpK_mathrma$ of $ceCaCl2$ 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 $mathrmpH$ probe to be basic. I've tried multiple sources of calcium chloride and each is basic.
TLDR:
Theoretically and literature values state that $ceCaCl2$ is acidic, but my empirical evidence shows that it is basic.
acid-base solutions ionic-compounds
edited May 11 at 6:26
Mathew Mahindaratne
7,7641030
asked May 10 at 12:53
124c41124c41
947
$endgroup$
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. $ceHCl$ is a strong acid, and so the salt should be slightly acidic. Wikipedia states the $mathrmpK_mathrma$ of $ceCaCl2$ 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 $mathrmpH$ probe to be basic. I've tried multiple sources of calcium chloride and each is basic.
TLDR:
Theoretically and literature values state that $ceCaCl2$ is acidic, but my empirical evidence shows that it is basic.
acid-base solutions ionic-compounds
acid-base solutions ionic-compounds
edited May 11 at 6:26
Mathew Mahindaratne
7,7641030
asked May 10 at 12:53
124c41124c41
947
edited May 11 at 6:26
Mathew Mahindaratne
7,7641030
asked May 10 at 12:53
124c41124c41
947
edited May 11 at 6:26
Mathew Mahindaratne
7,7641030
edited May 11 at 6:26
Mathew Mahindaratne
7,7641030
edited May 11 at 6:26
Mathew Mahindaratne
7,7641030
7,7641030
asked May 10 at 12:53
124c41124c41
947
asked May 10 at 12:53
124c41124c41
947
asked May 10 at 12:53
124c41124c41
947
947
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Calcium hydroxide has solubility about 1.9 g/L. This is enough to create pH above 11, i.e. a strongly basic solution.
$ceCaCl2$ solutions should be very slightly acidic if they were made from pure $ceCaCl2$. 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
answered May 10 at 14:02
permeakrapermeakra
18.6k13887
$endgroup$
$begingroup$
"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 at 17:30
1
$begingroup$
@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 at 20:48
$begingroup$
You mean : $ceCa(OH)2_mathrm(aq) <=>Ca(OH)^+_mathrm(aq) + OH^-_mathrm(aq)$ , so: $$[ceOH-]=[ceCa(OH)2] = frac1.974=pu0.025M$$
$endgroup$
– Adnan AL-Amleh
May 10 at 22:27
2
$begingroup$
@AdnanAL-Amleh Yes. 0.025 M for $[OH-] $ is a low estimate, it is probably higher because $ceCa(OH)+$ can dissociate too.
$endgroup$
– permeakra
May 10 at 22:55
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Calcium hydroxide has solubility about 1.9 g/L. This is enough to create pH above 11, i.e. a strongly basic solution.
$ceCaCl2$ solutions should be very slightly acidic if they were made from pure $ceCaCl2$. 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
answered May 10 at 14:02
permeakrapermeakra
18.6k13887
$endgroup$
$begingroup$
"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 at 17:30
1
$begingroup$
@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 at 20:48
$begingroup$
You mean : $ceCa(OH)2_mathrm(aq) <=>Ca(OH)^+_mathrm(aq) + OH^-_mathrm(aq)$ , so: $$[ceOH-]=[ceCa(OH)2] = frac1.974=pu0.025M$$
$endgroup$
– Adnan AL-Amleh
May 10 at 22:27
2
$begingroup$
@AdnanAL-Amleh Yes. 0.025 M for $[OH-] $ is a low estimate, it is probably higher because $ceCa(OH)+$ can dissociate too.
$endgroup$
– permeakra
May 10 at 22:55
add a comment |
$begingroup$
Calcium hydroxide has solubility about 1.9 g/L. This is enough to create pH above 11, i.e. a strongly basic solution.
$ceCaCl2$ solutions should be very slightly acidic if they were made from pure $ceCaCl2$. 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
answered May 10 at 14:02
permeakrapermeakra
18.6k13887
$endgroup$
$begingroup$
"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 at 17:30
1
$begingroup$
@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 at 20:48
$begingroup$
You mean : $ceCa(OH)2_mathrm(aq) <=>Ca(OH)^+_mathrm(aq) + OH^-_mathrm(aq)$ , so: $$[ceOH-]=[ceCa(OH)2] = frac1.974=pu0.025M$$
$endgroup$
– Adnan AL-Amleh
May 10 at 22:27
2
$begingroup$
@AdnanAL-Amleh Yes. 0.025 M for $[OH-] $ is a low estimate, it is probably higher because $ceCa(OH)+$ can dissociate too.
$endgroup$
– permeakra
May 10 at 22:55
add a comment |
$begingroup$
Calcium hydroxide has solubility about 1.9 g/L. This is enough to create pH above 11, i.e. a strongly basic solution.
$ceCaCl2$ solutions should be very slightly acidic if they were made from pure $ceCaCl2$. 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
answered May 10 at 14:02
permeakrapermeakra
18.6k13887
$endgroup$
Calcium hydroxide has solubility about 1.9 g/L. This is enough to create pH above 11, i.e. a strongly basic solution.
$ceCaCl2$ solutions should be very slightly acidic if they were made from pure $ceCaCl2$. 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
answered May 10 at 14:02
permeakrapermeakra
18.6k13887
edited May 10 at 20:49
answered May 10 at 14:02
permeakrapermeakra
18.6k13887
answered May 10 at 14:02
permeakrapermeakra
18.6k13887
answered May 10 at 14:02
permeakrapermeakra
18.6k13887
18.6k13887
$begingroup$
"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 at 17:30
1
$begingroup$
@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 at 20:48
$begingroup$
You mean : $ceCa(OH)2_mathrm(aq) <=>Ca(OH)^+_mathrm(aq) + OH^-_mathrm(aq)$ , so: $$[ceOH-]=[ceCa(OH)2] = frac1.974=pu0.025M$$
$endgroup$
– Adnan AL-Amleh
May 10 at 22:27
2
$begingroup$
@AdnanAL-Amleh Yes. 0.025 M for $[OH-] $ is a low estimate, it is probably higher because $ceCa(OH)+$ can dissociate too.
$endgroup$
– permeakra
May 10 at 22:55
add a comment |
$begingroup$
"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 at 17:30
1
$begingroup$
@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 at 20:48
$begingroup$
You mean : $ceCa(OH)2_mathrm(aq) <=>Ca(OH)^+_mathrm(aq) + OH^-_mathrm(aq)$ , so: $$[ceOH-]=[ceCa(OH)2] = frac1.974=pu0.025M$$
$endgroup$
– Adnan AL-Amleh
May 10 at 22:27
2
$begingroup$
@AdnanAL-Amleh Yes. 0.025 M for $[OH-] $ is a low estimate, it is probably higher because $ceCa(OH)+$ can dissociate too.
$endgroup$
– permeakra
May 10 at 22:55
$begingroup$
"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 at 17:30
$begingroup$
"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 at 17:30
1
1
$begingroup$
@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 at 20:48
$begingroup$
@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 at 20:48
$begingroup$
You mean : $ceCa(OH)2_mathrm(aq) <=>Ca(OH)^+_mathrm(aq) + OH^-_mathrm(aq)$ , so: $$[ceOH-]=[ceCa(OH)2] = frac1.974=pu0.025M$$
$endgroup$
– Adnan AL-Amleh
May 10 at 22:27
$begingroup$
You mean : $ceCa(OH)2_mathrm(aq) <=>Ca(OH)^+_mathrm(aq) + OH^-_mathrm(aq)$ , so: $$[ceOH-]=[ceCa(OH)2] = frac1.974=pu0.025M$$
$endgroup$
– Adnan AL-Amleh
May 10 at 22:27
2
2
$begingroup$
@AdnanAL-Amleh Yes. 0.025 M for $[OH-] $ is a low estimate, it is probably higher because $ceCa(OH)+$ can dissociate too.
$endgroup$
– permeakra
May 10 at 22:55
$begingroup$
@AdnanAL-Amleh Yes. 0.025 M for $[OH-] $ is a low estimate, it is probably higher because $ceCa(OH)+$ can dissociate too.
$endgroup$
– permeakra
May 10 at 22:55
add a comment |
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