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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: ed_saifa on March 09, 2008, 06:50:29 pm

Title: please help with chemistry questions
Post by: ed_saifa on March 09, 2008, 06:50:29 pm: Hey
Question:
Anhydrous sodium carbonate was used in an experiment to determine the concentration of an unknown hydrochloric acid solution. A sample of the carbonate was heated, cooled and weighed until constant weight was obtained.

(a)After being treated as described above, 1.377g of anhydrous sodium carbonate was dissolved in water and made up to 250cm ^3 in a volumetric flask. A 25.00ml portion of this solution was titrated with hydrochloric acid, using methyl orange as indicator. The titration required 22.65mL of the acid for complete neutralization.
(i) Determine the molar concentration of the sodium carbonate solution.
(ii) Determine the molar concentration of the unknown hydrochloric acid solution

(b) A second solution was prepared by dissolving 0.5g of crystalline sodium carbonate in water. This solution was then titrated with another portion of the hydrochloric acid solution. It was found that 30.45mL of the acid was required for complete neutralization.
(i) Calculate the apparent molar mass of the sodium carbonate used in the procedure
(ii) Determine the molecular formula of the crystals used in this procedure

Thank you!
Title: Re: please help with chemistry questions
Post by: Collin Li on March 09, 2008, 07:22:05 pm: a)

i.

$\mbox{[Na}_2\mbox{CO}_3\mbox{]} = \frac{n}{V} = \frac{1.377\mbox{ g}}{106\mbox{ gmol}^{-1}} \times \frac{1}{0.250\mbox{ L}} = 0.0520\mbox{ M}$

ii.

$\mbox{n(Na}_2\mbox{CO}_3\mbox{)} = 0.02500 \times 0.0520 = 0.00130\mbox{ mol}$

Chemical equation: $\mbox{2HCl + Na}_2\mbox{CO}_3 \rightarrow \mbox{2NaCl + H}_2\mbox{O + CO}_2$

$\implies \mbox{n(HCl)} = 2\cdot \mbox{n(Na}_2\mbox{CO}_3\mbox{)} = 0.00260\mbox{ mol}$

$\implies \mbox{[HCl]} = \frac{n}{V} = \frac{0.00260}{0.02265} = 0.115\mbox{ M}$
Title: Re: please help with chemistry questions
Post by: Toothpaste on March 09, 2008, 07:30:20 pm: $Na_2CO_3 (aq) + 2HCl (aq) \rightarrow 2NaCl (aq) + CO_2 (g) + H_2O (l)$

$M_r(Na_2CO_3) = 106gmol^{-1}$
$m(Na_2CO_3) = 1.377g$

$n(Na_2CO_3)= \frac{1.377}{106} = 0.01299mol$

$c(Na_2CO_3) = \frac{0.01299mol}{0.250L} = 0.05196226 M = 0.0520M$

Find the moles of $Na_2CO_3$ in the titration
given: $V = 25mL$ and $c = 0.05196226 M$

$n = cV = 0.001299mol$

Using the equation
$n(HCl) = 2 * n(Na_2CO_3) = 0.002598mol$
$c(HCl) = 0.1147 M = 0.115 M$
Title: Re: please help with chemistry questions
Post by: ed_saifa on March 09, 2008, 07:31:45 pm: thanks!
Title: Re: please help with chemistry questions
Post by: Collin Li on March 09, 2008, 07:32:43 pm: Toothpick, you're right. I have fixed this consequential error. I realised this when doing part b.

Thanks!
Title: Re: please help with chemistry questions
Post by: Collin Li on March 09, 2008, 07:37:25 pm: b)

Realise that crystalline sodium carbonate refers to hydrated sodium carbonate (is of the form $\mbox{Na}_2\mbox{CO}_3\cdot x\mbox{H}_2\mbox{O}$ ).

i.

$\mbox{n(HCl)} = 0.115\times 0.03045 = 0.00349\mbox{ mol}$

Since the stoichometric ratio between anhydrous sodium carbonate and hydrochloric acid should be equivalent to the reaction with the hydrated form (because the additional waters do not change the acid-base reaction), we can say:

$\mbox{n(Na}_2\mbox{CO}_3\cdot x\mbox{H}_2\mbox{O)} = \frac{\mbox{n(HCl)}}{2} = 0.00175\mbox{ mol}$

$M = \frac{m}{n} = \frac{0.5}{0.00175} = 286$

ii.

Using $M = 106 + 18x$ (dry sodium carbonate + an unknown amount of water):

$\implies 106+18x = 286$

$\implies x = 10$

Therefore, the molecular formula is:
$\mbox{Na}_2\mbox{CO}_3\cdot 10\mbox{H}_2\mbox{O}$
Title: Re: please help with chemistry questions
Post by: ed_saifa on March 09, 2008, 08:23:27 pm: Thanks heaps
Title: Re: please help with chemistry questions
Post by: ed_saifa on March 10, 2008, 08:16:51 pm: Hi, i've got another questions. Help would be hot! XD Thanks in advance
Question 5.
An aqueous solution of potassium dichromate, K2Cr2O7, was prepared by dissolving 2.31 g of the pure solid in enough water to make up 500 mL of solution.
A piece of iron wire, weighing 0.597 g was added to a dilute acid solution so that all the Fe(s) atoms were converted to Fe2+(aq) ions. The resulting solution was diluted to 100 mL.
25.0 mL aliquots of the Fe2+(aq) solution were further acidified and titrated with the potassium dichromate solution.
During the titration Fe2+(aq) ions were oxidised to Fe3+(aq) ions by the Cr2O72-(aq) ions, which were converted to Cr3+(aq) ions.
The average titre was 27.3 mL
(a)   Write a balanced equation for the conversion of Fe(s) to Fe2+(aq) in acidic solution
(b)   The titration involves a redox reaction. Write (I) the oxidation half-equation, (ii) the reduction half-equation, and (iii) the overall redox equation describing this reaction.
(c)   Determine the concentration of the potassium dichromate solution.
(d)   Determine the percentage purity of the iron wire.
Title: Re: please help with chemistry questions
Post by: Collin Li on March 10, 2008, 08:44:33 pm: Here's the first few parts (the theory parts)

For the redox equations with iron, just use commonsense (balance the atoms on both sides) and/or refer to the electrochemical series:

a) $\mbox{Fe}_{\mbox{(s)}}\mbox{ + 2H}^+\,_{\mbox{(aq)}} \rightarrow \mbox{Fe}^{2+}\,_{\mbox{(aq)}}\mbox{ + H}_2_{\mbox{(g)}}$

b)

i) oxidation: $\mbox{Fe}^{2+}\,_{\mbox{(aq)}} \rightarrow \mbox{Fe}^{3+}\,_{\mbox{(aq)}}\mbox{ + e}^-$

ii) This one we have to work out: it goes from $\mbox{Cr}_2\mbox{O}_7\,^{2-}\,_{\mbox{(aq)}}$ to $\mbox{Cr}^{3+}\,_{\mbox{(aq)}}$ so we use those balancing rules for redox (balance oxygens with waters, then hydrogen atoms with $\mbox{H}^+$ , and charges with electrons). You should get:

reduction: $\mbox{Cr}_2\mbox{O}_7\,^{2-}\,_{\mbox{(aq)}}\mbox{ + 14H}^+\,_{\mbox{(aq)}}\mbox{ + 6e}^- \rightarrow \mbox{2Cr}^{3+}\,_{\mbox{(aq)}}\mbox{ + 7H}_2\mbox{O}_{\mbox{(l)}}$

iii) Adding 6 times the oxidation reaction and 1 times the reduction reaction (to balance the number of electrons) yields:

overall reaction: $\mbox{6Fe}^{2+}\,_{\mbox{(aq)}}\mbox{ + Cr}_2\mbox{O}_7\,^{2-}\,_{\mbox{(aq)}}\mbox{ + 14H}^+\,_{\mbox{(aq)}} \rightarrow \mbox{6Fe}^{3+}\,_{\mbox{(aq)}}\mbox{ + 2Cr}^{3+}\,_{\mbox{(aq)}}\mbox{ + 7H}_2\mbox{O}_{\mbox{(l)}}$
Title: Re: please help with chemistry questions
Post by: Collin Li on March 10, 2008, 08:56:04 pm: c) $\mbox{[K}_2\mbox{Cr}_2\mbox{O}_7\mbox{]} = \frac{2.31\mbox{ g}}{294.2\mbox{ gmol}^{-1}}\times \frac{1}{0.500\mbox{ L}} = 0.0157\mbox{ M}$

d) Find the amount of potassium dichromate consumed in reaction with iron:

$\mbox{n(K}_2\mbox{Cr}_2\mbox{O}_7\mbox{)} = 0.0157\times 0.0273 = 4.29\times 10^{-4}\mbox{ mol}$

$\mbox{n(Fe)}_{\mbox{aliquot}} = \mbox{n(Fe}^{2+}\mbox{)} = 6\cdot\mbox{n(K}_2\mbox{Cr}_2\mbox{O}_7\mbox{)} = 0.00257\mbox{ mol}$

Recognise that this is only the number of moles of iron in one aliquot (25.0 mL):

$\implies \mbox{n(Fe)} = \mbox{n(Fe)}_{\mbox{aliquot}} \times \frac{100}{25.0} = 0.0103\mbox{ mol}$

$\implies \mbox{m(Fe)} = 0.0103 \times 55.9 = 0.575\mbox{ g}$

$\implies \mbox{\�} = \frac{0.575}{0.597}\times 100 = 96.3\mbox{\% (w/w)}$
Title: Re: please help with chemistry questions
Post by: ed_saifa on March 10, 2008, 09:08:11 pm: Thanks heaps Coblin!!
Title: Re: please help with chemistry questions
Post by: ed_saifa on March 11, 2008, 08:49:02 pm: Could anyone please help me with this question?

One molecule of vitamin B contains 63 carbon atoms. The percentage by mass of carbon in Vitamin B is 55.74 %. What is the molar mass of vitamin B?

Thanks
Title: Re: please help with chemistry questions
Post by: midas_touch on March 11, 2008, 09:13:35 pm: Mass of 63 carbon atoms: 63*12 = 756 amu.

Percentage of carbon in Vitamin B = 55.74%

Hence, molar mass of Vitamin B is given by
756/(0.5574) g/mol
= 1356.3 g/mol

Keeping in mind that 1 amu is exactly equal to 1 g/mol.
Title: Re: please help with chemistry questions
Post by: ed_saifa on March 11, 2008, 09:16:47 pm: thanks heaps midas_touch! i would raise your karma but i dont know how =(
Title: Re: please help with chemistry questions
Post by: Toothpaste on March 11, 2008, 09:26:11 pm: Quote from: ed_saifa on March 11, 2008, 09:16:47 pm
thanks heaps midas_touch! i would raise your karma but i dont know how =(

You need 50 posts to do that. :)
Title: Re: please help with chemistry questions
Post by: ed_saifa on March 11, 2008, 09:31:01 pm: oh well! i guess you learn something new everyday! thanks for that Toothpick XD