Homework Help

What volume, in mL, of 0.1234 M HCl is needed to neutralize 2.00 g Ca(OH)2?

clairelizz's profile pic

Posted via web

dislike 1 like

What volume, in mL, of 0.1234 M HCl is needed to neutralize 2.00 g Ca(OH)2?

2 Answers | Add Yours

jerichorayel's profile pic

Posted (Answer #1)

dislike 1 like

First let us make a balanced chemical equation for this reaction.

`2 HCl + Ca(OH)_2 -> CaCl_2 + 2 H_2O`

To solve this problem, we should first compute how many moles of Ca(OH)2 are needed to be reacted. Using the above equation, we can determine the amount of moles of HCl and eventually the volume of the HCl solution.

`2.00 grams Ca(OH)_2 * (1 mol e Ca(OH)_2)/(74.093 grams Ca(OH)_2) * (2 mol es HCl)/(1 mol e Ca(OH)_2)`

= 0.05399 moles HCl

From molarity:

`Molarity = (mol es solute)/(volume of solution (L))`

`Volume of solution = (0.05399 mol es HCl)/(0.1234 (mol es)/(L))`

Volume of solution = 0.4375 L = 438 mL

 

Note: 1000mL = 1L

Sources:

alishashah's profile pic

Posted (Answer #2)

dislike 0 like

The concept of gram equivalents is helpful in such questions.

1 gmeq= 1000 meq (meq= milliequivalent)

and meq = molarity X n-factor X volume(in mL) (for aqueous solution of acids and bases)

meq = given mass(in gms) X 1000 X n-factor/molar mass (for solids)

meq of acid(HCl) = meq of base (Ca(OH)2)

=> 0.1234 X 1 X v(ml)= 1000 X 2/ 74 X 2

=> V(ml)= 438.039 mL

I hope this helped!! You may refer to the links given for more information on the concept of gram-equivalents 

 

 

 

 

 

 

 

Sources:

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes