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# Chemistry Back-Titration Question

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1. I am getting a stupid answer for this question. Can someone tell me where I've gone wrong or talk me through the right answer? Thanks

Question:
A sample of chalk weighing 0.135g was reacted with 40.00cm3 of 1mol/l hydrochloric acid (an excess). After the reaction was complete the solution was transferred to a 250cm3 volumetric flask and made up to the mark with distilled water. 25.00cm3 portions of this solution were then titrated against 0.15mol/l sodium hydroxide using methyl orange as the indicator. An average of 25cm3 of the sodium hydroxide was required.
Calculate the percentage purity of the chalk.

HCl + NaOH --> H2O + NaCl
1 mol HCl reacts with 1 mol NaOH
No. moles NaOH = 0.15 x 0.025 = 0.00375mol
0.00375 mol HCl reacts with 0.00375 mol NaOH

0.00375 mol in 1/10 of reacted solution
0.0375 mol in reacted solution
No. moles HCl = 0.04 x 1 = 0.04 mol
No. moles HCl reacted = 0.04 - 0.037 = 0.003 mol

CaCO3 + 2HCl --> CaCl2 + CO2 + H2O
1 mol CaCO3 reacts with 2 mol HCl
0.0015 mol reacts with 0.003 mol HCl

1 mol CaCO3 = 100g
0.0015 mol = 0.15g

This would give 0.15/0.135x100 which is 111.11% purity. This is stupid
2. Hi,
When you did the subtraction for the moles of HCL, you left out a decimal place. put it back in and there are:
0.0025moles of HCL reacted
/2 =0.00125moles of CaCO3 on the 0.135g sample
=0.125g of calcium carbonate in the sample
% purity = 0.125/0.135 x 100
=92.6% pure (3sf)
Hope that helps!
3. (Original post by chris-carnage)
I am getting a stupid answer for this question. Can someone tell me where I've gone wrong or talk me through the right answer? Thanks

Question:
A sample of chalk weighing 0.135g was reacted with 40.00cm3 of 1mol/l hydrochloric acid (an excess). After the reaction was complete the solution was transferred to a 250cm3 volumetric flask and made up to the mark with distilled water. 25.00cm3 portions of this solution were then titrated against 0.15mol/l sodium hydroxide using methyl orange as the indicator. An average of 25cm3 of the sodium hydroxide was required.
Calculate the percentage purity of the chalk.

Total moles of HCl = 0.04 x 1 = 0.04 mol
Moles of NaOH in the titre = 0.025 x 0.15 = 0.00375
Therefore moles of NaOH requiired to neutralise all of the excess acid = 0.0375 mol
Acid used up in the reaction = 0.04 - 0.0375 = 0.0025 mol

CaCO3 + 2HCL = products

Mol of CaCO3 = 0.00125 (Mr = 100)

Mass = 0.125g
percent purity = 0.125/0.135 * 100 = 92.6%

HCl + NaOH --> H2O + NaCl
1 mol HCl reacts with 1 mol NaOH
No. moles NaOH = 0.15 x 0.025 = 0.00375mol
0.00375 mol HCl reacts with 0.00375 mol NaOH

0.00375 mol in 1/10 of reacted solution
0.0375 mol in reacted solution
No. moles HCl = 0.04 x 1 = 0.04 mol
No. moles HCl reacted = 0.04 - 0.037 = 0.003 mol (here you have rounded the 0.0375)

CaCO3 + 2HCl --> CaCl2 + CO2 + H2O
1 mol CaCO3 reacts with 2 mol HCl
0.0015 mol reacts with 0.003 mol HCl

1 mol CaCO3 = 100g
0.0015 mol = 0.15g

This would give 0.15/0.135x100 which is 111.11% purity. This is stupid
4. (Original post by charco)

Total moles of HCl = 0.04 x 1 = 0.04 mol
Moles of NaOH in the titre = 0.025 x 0.15 = 0.00375
Therefore moles of NaOH requiired to neutralise all of the excess acid = 0.0375 mol
Acid used up in the reaction = 0.04 - 0.0375 = 0.0025 mol

CaCO3 + 2HCL = products

Mol of CaCO3 = 0.00125 (Mr = 100)

Mass = 0.125g
percent purity = 0.125/0.135 * 100 = 92.6%
Thank you Didn't realise rounding could cause such a problem! (I probably should though!)
Updated: April 5, 2010
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