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