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A 15.00 mL sample of HCl solution of unknown concentration was titrated with .185 M...

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hardguy | eNotes Newbie

Posted September 2, 2013 at 5:59 AM via web

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A 15.00 mL sample of HCl solution of unknown concentration was titrated with .185 M Ca(OH)2. It took 23.9 mL of the base to reach the equivalence point of the titration. What is the concentration of the acid?

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted September 2, 2013 at 6:05 AM (Answer #1)

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`Ca(OH)_2+2HCl rarr CaCl_2+2H_2O`

Mole ratio

`Ca(OH)_2:HCl = 1:2`

Amount of `Ca(OH)_2` consumed `= 0.185/1000xx23.9 = 0.00442`

Amount of HCl moles reacted `= 2xx0.00442`

Let us say the concentration of the HCl solution is M.

`M/1000xx15 = 2xx0.00442`

`M = 0.5893`

So the concentration of the unknown HCl solution is 0.5893M.

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ayl0124 | TA , Grade 12 | Valedictorian

Posted August 13, 2014 at 3:12 AM (Answer #2)

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`"M_AV_A=M_BV_B"`

At the equivalence point, the moles of H+ must equal the moles of OH-.

"A" is for acid and "B" is for base:

`"M_A"(15"mL")=(0.185"M")(23.9"mL")`

Solve for the molarity of H+.

`"M_A" = 0.2947`

Notice that the mole ratio is different between HCl and Ca(OH)2. You must double the moles of H+ to account for this unbalanced ratio. 

`2(0.2947) = 0.590"M"`

The initial concentration of HCl is 0.590 M. Note that there needs to be three sig figs. 

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