Titration Q. (1 Viewer)

Da_Bomb · Apr 19, 2010

how to work out this q?.

5.267 g anhydrous sodium carbonate was dissolved in water in a volumetric flask and the volume made up to 250ml. 10ml of this solution was pippeted into a conical flask and titrated with hydrochloric acid. 21.3 ml was needed to reach the equivalence point. Calculate the molarity of the hcl solution.

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[P.s. Answer: 0.187 mol / L ]

adomad · Apr 21, 2010

$first find the concentration of the sodium carbonate solution you made$
$c_{Na_2CO_3} =\frac{moles}{volume}$
$$then write a balanced equation of the reaction between the Sodium Carbonate and the hydrochloric acid%$
$Na_2CO_3 + HCl \to 2NaCl + H_2O + CO_2$
$now work out the moles that is in that 10 mLs that you have pippetted and placed in the conical flask, by using:$
$c=\frac{n}{v} $ we calculate the concentration in the first line and we know the volume = 0.01 L... solve for moles$
$now from the balanced equation, find the moles of HCl that will react with the number of moles that you found in the 10 mL sample of the Na_2CO_3 $ $ solution$
$now that we know the moles of the HCl needed, use the formula:$
$c=\frac{n_{HCl}}{v}, $ where the volume in this case is 21.3 mLs$
$NOTE: remember to convert the mLs into L... standart SI Units :D$

Hope this helps

liveyourlife · May 19, 2010

its not balanced is it?
isnt it meant to be
na2co3 + 2hcl -> h20 + co2 + 2nacl

Titration Q. (1 Viewer)

Da_Bomb

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adomad

HSC!!

liveyourlife

New Member

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