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according to the balanced equation what is the answer to the problem

2HCl + Ca(OH)2 --> CaCl2 + 2H2O

How many mL of 4.7 M HCl are required to neutralize (react completely) with a 868.7 mL of 4.69 M Ca(OH)2?

 

i need some help please explain your work

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You are  confused ?

Well so am  I.

Calcium hydroxide is  barely soluble with a solubility product  of 5.5 x 10-6

So I make its solubility as 0.0112M at  room temperature.

 

Did  you copy the molarities down correctly?

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5 minutes ago, studiot said:

You are  confused ?

Well so am  I.

Calcium hydroxide is  barely soluble with a solubility product  of 5.5 x 10-6

So I make its solubility as 0.0112M at  room temperature.

 

Did  you copy the molarities down correctly?

Could it be a trick question?

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It would not be a true solution; it would be heterogeneous.  It could be that the problem is imperfect, as opposed to being a trick question.  As someone who has written my fair share of homework problems, I can say that it easy to overlook difficulties like this.  Nevertheless, the amount of HCl needed to neutralize it is still something that could be calculated.

Edited by BabcockHall
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52 minutes ago, BabcockHall said:

It would not be a true solution; it would be heterogeneous.  It could be that the problem is imperfect, as opposed to being a trick question.  As someone who has written my fair share of homework problems, I can say that it easy to overlook difficulties like this.  Nevertheless, the amount of HCl needed to neutralize it is still something that could be calculated.

There would be a 0.01M solution, with a quickly settling sludge in the bottom of the container.

'Imperfect' is a polite way of  describing such a problem.

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38 minutes ago, John Cuthber said:

Did anybody specify that the solvent is water?
I must admit, I can't easily think of anything in which you could prepare a 4.69M solution of Ca(OH)2

Wouldn't a different solvent also have to dissolve HCl ?

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5 hours ago, studiot said:

Wouldn't a different solvent also have to dissolve HCl ?

There would be nothing to stop you using water as the solvent for the acid and a different solvent for the base- as long as the solvents mix (and no awkward side reactions happen).
Maybe glycerin or glycol would dissolve enough calcium hydroxide; they are certainly better solvents than water is.

 

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3 hours ago, John Cuthber said:

There would be nothing to stop you using water as the solvent for the acid and a different solvent for the base- as long as the solvents mix (and no awkward side reactions happen).

That's true enough.

Quote

https://pubchem.ncbi.nlm.nih.gov/compound/Calcium-hydroxide#section=Solubility

0.2 % at 32° F (NIOSH, 2016)

National Institute of Occupational Safety and Health. NIOSH Pocket Guide to Chemical Hazards (full website version). https://www.cdc.gov/niosh/npg (accessed August 2016).
 

Slightly soluble in water. Insoluble in ethanol. Soluble in glycerol

 

In water, 1730 mg/L at 20 °C

Shiu WY et al; Rev Environ Contam Toxicol 116: 15-187 (1990)
 

In water, 0.160 g/100 g water at 20 °C

Haynes, W.M. (ed.). CRC Handbook of Chemistry and Physics. 94th Edition. CRC Press LLC, Boca Raton: FL 2013-2014, p. 4-55
 

Very slightly soluble in boiling water. Insoluble in alcohol.

O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Cambridge, UK: Royal Society of Chemistry, 2013., p. 292
 

Soluble in acid

Haynes, W.M. (ed.). CRC Handbook of Chemistry and Physics. 94th Edition. CRC Press LLC, Boca Raton: FL 2013-2014, p. 4-55
 

Soluble in glycerol, sugar or ammonium chloride solution; soluble in acids with evolution of much heat

O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Cambridge, UK: Royal Society of Chemistry, 2013., p. 292
 

Solubility in water: none

 

(32°F): 0.2%

 

 

Edited by studiot
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