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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: thomas49th on April 04, 2007, 02:41:00 PM

Title: Check my titration
Post by: thomas49th on April 04, 2007, 02:41:00 PM

Hi is the answer to this question

50 cm3 of 2 mol dm-3 NaOH neutralises 30 cm3 of what concentration HCl?

3.33 mol/dm³

I first wrote out the balanced equation
NaOH + HCl -> NaCl + H₂O

So the moles are equal

I then calculated the mol for NaOH which came out as 0.1

I then divided 0.1 by 0.03 (which is 30cm³  converted to dm³)

This gave me 3.3333 recurring which I rounded to 2dp.

Is that right?

Thanks

Title: Re: Check my titration
Post by: Farid on April 04, 2007, 02:54:19 PM

yEs. Your result is perfect :)

Title: Re: Check my titration
Post by: DevaDevil on April 04, 2007, 03:01:39 PM

yes, the answer is 3 mol dm^-3; mind the significant figures though.

You start with 50 cm^-3 of 2 M NaOH (not 2.0), hence only 1 significant figure.