This is a past paper question taken from 2008 HKALE Chemistry Paper 2 Section A Question 3a.
(i) Sketch the variation of pH when 25.0 cm^3 of 0.05 M Na2CO3 (aq) is titration against 0.1 M HCl (aq).
The answer has got two sudden drop in pH. I don't understand.
(ii) A solid sample was known to contain Na2CO3 (s), NaHCO3 (s) and other water-soluble inert substances. 2.382 g of the sample was dissolved in deionised water and made up to 250.0 cm^3. 25.00 cm^3 of the solution was withdrawn and titrated against 0.0980 M HCl (aq) using phenolphthalein as an indicator. When 13.4 cm^3 of the HCl (aq) was added, the solution instantaneously changed from pink to colourless. A few drops of methyl organge was then added to the resulting solution and the titration was continued. Addition of further 23.65 cm^3 of the HCl (aq) caused the solution to turn orange.
Calculate the percentage by mass of Na2CO3 (s) and that of NaHCO3 (s).
I have read the marking scheme but I don't understand the steps. I also do not know what the word 'further' mean. Does it mean add 23.65 cm^3 of HCl or (23.65-13.4) cm^3 of HCl?
Thank you for your answer.
Question on titration
2010-12-31 1:27 am
回答 (1)
2010-12-31 4:53 am
✔ 最佳答案
(i)Before addition of HCl, Na2CO3 is rather alkaline due to the hydrolysis of CO32- ions giving OH⁻ ions.
The titration consists of two stages:
The 1st stage:
Na2CO3(aq) + HCl(aq) → NaHCO3(aq) + H2O(l)
At the 1st end point, the solution of HCO3⁻ ions are slightly alkaline.
The 2nd stage:
NaHCO3(aq) + HCl(aq) → NaCl(aq) + H2O(l) + CO2(aq)
At the 2nd end point, the CO2 formed is slightly soluble in water to give a slightly acidic solution.
The two sudden drops in pH are corresponding to the two end points.
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(ii)
Let a mol be the number of moles of Na2CO3 in the sample, and b mol be that of NaHCO3.
No. of moles of Na2CO3 used in titration = a x (25/250) = 0.1a mol
No. of moles of NaHCO3 used in titration = b x (25/250) = 0.1b mol
When using phenolphthalein as an indicator, only Na2CO3 is reacted to give NaHCO3.
Na2CO3(aq) + HCl(aq) → NaHCO3(aq) + H2O(l)
No. of moles of HCl used = 0.098 x (13.4/1000) = 0.001313 mol
No. of moles of Na2CO3 in the sample = 0.1a mol = 0.001313 mol
Hence, a = 0.01313
No. of moles of NaHCO3 formed in the above reaction = 0.001313 mol
When using methyl orange as an indicator, NaHCO3 (original + formed in the above reaction) reacts with HCl to give CO2.
NaHCO3(aq) + HCl(aq) → NaCl(aq) + H2O(l) + CO2(aq)
("Further 23.65 cm³ of the HCl" means that after the addition of 13.4 cm³ of the HCl and methyl orange indicator, 23.65 cm³ more of the HCl is then added.)
In the above reaction:
No. of moles of HCl used = 0.098 x (23.65/1000) = 0.002318 mol
No. of moles of Na2CO3 used = (0.001313 + 0.1b) mol = 0.002318 mol
0.1b = 0.001005
b = 0.01005
Molar mass of Na2CO3 = 22.99x2 + 12.01 + 16x3 = 105.99 g mol⁻¹
Mass of Na2CO3 in the sample = 0.01313 x 105.99 = 1.392 g
Molar mass of NaHCO3 = 22.99 + 1.008 + 12.01 + 16x3 = 84.008
Mass of NaHCO3 in the sample = 0.01005 x 84.008 = 0.8443 g
% by mass of Na2CO3 = (1.392/2.382) x 100% = 58.44%
% by mass of NaHCO3 = (0.8443/2.382) x 100% = 35.45%
2010-12-31 02:19:37 補充：
But I don't understand why will Na2CO3 react first before NaHCO3. Has the question implied this?
This titration is included in the syllabus, you have to know this. Otherwise, you can get in by interpretation of the graph.
2010-12-31 02:24:31 補充：
The answer says that it should be '(23.65-13.4)/1000' instead of No. of moles of HCl used = 0.098 x (23.65/1000) = 0.002318 mol
2010-12-31 02:24:44 補充：
No. of moles of HCl reacted with total NaHCO3 = 0.098 x (23.65/1000)
No. of moles of HCl reacted with NaHCO3 from Na2CO3 = 0.098 x (13.4/1000)
No. of moles of HCl reacted with original NaHCO3 = 0.098 x (23.65 - 13.4)/1000
參考: micatkie
收錄日期: 2021-04-13 17:44:11
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