Is it me or there a mistake somewhere here??

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XShmalX
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#1
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#1
An impure sample of barium hydroxide of mass 1.6524g was allowed to react with 100cm3 of hydrochloric acid of concentration 0.200 M. When the excess acid was titrated against sodium hydroxide, 10.9 cm3 of the sodium hydroxide solution required. 28.5 cm3 of the same hydrochloric acid for neutralization in a separate titration. Calculate the percentage of purity of the sample of barium hydroxide.


Moles HCl = 0.0285 L x 0.200 = 0.00570
= moles NaOH
concentration NaOH = 0.00570 mol / 0.0109 L = 0.523 M
Moles HCl used in the titration = 0.200 x 0.100 = 0.0200
moles HCl = 0.0200 - 0.00570 =0.0143

Ba(OH)2 + 2 HCl >> BaCl2 + 2 H2O
moles Ba(OH)2 = 0.0143 / 2 =0.00715
Mass Ba(OH)2 = 0.00715 mol x 171.342 g/mol =1.225 g
% = 1.225 x 100 / 1.6524 =74.14
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charco
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#2
Report 12 years ago
#2
(Original post by XShmalX)
An impure sample of barium hydroxide of mass 1.6524g was allowed to react with 100cm3 of hydrochloric acid of concentration 0.200 M. When the excess acid was titrated against sodium hydroxide, 10.9 cm3 of the sodium hydroxide solution required 28.5 cm3 of the same hydrochloric acid for neutralization in a separate titration. Calculate the percentage of purity of the sample of barium hydroxide.


Moles HCl = 0.0285 L x 0.200 = 0.00570
= moles NaOH
concentration NaOH = 0.00570 mol / 0.0109 L = 0.523 M
Moles HCl used in the titration = 0.200 x 0.100 = 0.0200
moles HCl = 0.0200 - 0.00570 =0.0143

Ba(OH)2 + 2 HCl >> BaCl2 + 2 H2O
moles Ba(OH)2 = 0.0143 / 2 =0.00715
Mass Ba(OH)2 = 0.00715 mol x 171.342 g/mol =1.225 g
% = 1.225 x 100 / 1.6524 =74.14
It's a horribly convoluted question...
10.9 cm3 of the sodium hydroxide solution required 28.5 cm3 of the same hydrochloric acid for neutralization in a separate titration
what does this actually mean? The excess HCl or the original HCl? It says a separate titration so presumably its a reaction between NaOH and the original HCl - so this can be used to find the molarity of the NaOH...

but if that is the case how do you find the molarity of the excess acid without another titration result?

The question lacks data (and sense)
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XShmalX
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#3
Report Thread starter 12 years ago
#3
(Original post by charco)
It's a horribly convoluted question...


what does this actually mean? The excess HCl or the original HCl? It says a separate titration so presumably its a reaction between NaOH and the original HCl - so this can be used to find the molarity of the NaOH...

but if that is the case how do you find the molarity of the excess acid without another titration result?

The question lacks data (and sense)
Sorry I missed out a full stop after required oppps! :o: I'll change that!
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charco
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#4
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#4
(Original post by XShmalX)
Sorry I missed out a full stop after required oppps! :o: I'll change that!
An impure sample of barium hydroxide of mass 1.6524g was allowed to react with 100cm3 of hydrochloric acid of concentration 0.200 M. When the excess acid was titrated against sodium hydroxide, 10.9 cm3 of the sodium hydroxide solution required.

28.5 cm3 of the same hydrochloric acid for neutralization in a separate titration. - This is not a sentence. There is no verb!

Calculate the percentage of purity of the sample of barium hydroxide

So, if I understand your shorthand, you first determine the concentration of the NaOH (the red bit)...

But as you are not given the volume of the NaOH it is not possible.

If you want help I suggest that you type out the question correctly.
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XShmalX
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#5
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#5
(Original post by charco)
An impure sample of barium hydroxide of mass 1.6524g was allowed to react with 100cm3 of hydrochloric acid of concentration 0.200 M. When the excess acid was titrated against sodium hydroxide, 10.9 cm3 of the sodium hydroxide solution required.

28.5 cm3 of the same hydrochloric acid for neutralization in a separate titration. - This is not a sentence. There is no verb!

Calculate the percentage of purity of the sample of barium hydroxide

So, if I understand your shorthand, you first determine the concentration of the NaOH (the red bit)...

But as you are not given the volume of the NaOH it is not possible.

If you want help I suggest that you type out the question correctly.
Ahh sorry :o: Should be:
12) An impure sample of barium hydroxide of mass 1.6524 g was allowed to react with 100 cm3 of 0.200 mol dm-3
hydrochloric acid. When the excess acid was titrated against sodium hydroxide, 10.9 cm3 of sodium
hydroxide solution was required. 25.0 cm3 of the sodium hydroxide required 28.5 cm3 of the hydrochloric acid
in a separate titration. Calculate the percentage purity of the sample of barium hydroxide.
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XShmalX
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#6
Report Thread starter 12 years ago
#6
Really sorry!
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charco
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#7
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#7
(Original post by XShmalX)
Ahh sorry :o: Should be:
12) An impure sample of barium hydroxide of mass 1.6524 g was allowed to react with 100 cm3 of 0.200 mol dm-3
hydrochloric acid. When the excess acid was titrated against sodium hydroxide, 10.9 cm3 of sodium
hydroxide solution was required. 25.0 cm3 of the sodium hydroxide required 28.5 cm3 of the hydrochloric acid
in a separate titration. Calculate the percentage purity of the sample of barium hydroxide.

You still have not said how much of the excess acid was titrated against the NaOH - all of it?

OK lets assume all of it..


25.0 cm3 of the sodium hydroxide required 28.5 cm3 of the hydrochloric acid

moles acid = moles base = 0.0285 x 0.2 = 0.0057
Thus molarity of NaOH = 0.0057/0.025 = 0.228 M

When this titrated the excess acid 10.9 ml was used...
moles of excess acid = 0.0109 x 0.228 = 0.002485

Initial moles of acid = 0.1 x 0.2 = 0.02

therefore moles acid reacted = 0.02 - 0.002485 = 0.0175 mol

HCl reacts with Ba(OH)2 in a 2 :1 ratio

2HCl + Ba(OH)2 --> BaCl2 + 2H2O

thus moles of Ba(OH)2 = 0.0175/2 = 0.00876

mass = moles x RMM = 0.00876 x 171 = 1.4975 g

original mass weighed 1.6524 so percentage purity = 1.4975/1.6524 = 90.6%
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xell
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#8
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#8
^ Answer above may be incorrect as periodic table shows Barium is 137.3 where they've incorrectly used 137
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