- Level: International Baccalaureate
- Subject: Chemistry
- Word count: 1635
Titration of Na2CO3.xH2O with HCl
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Introduction
Chemistry Lab Report - Titration Introduction: Aim: To find out the number of moles of the water of crystallization (x) in hydrated sodium carbonate (Na2CO3.xH2O), by titrating it with hydrochloric acid (HCl). Hypothesis: As HCl is continuously added to the sodium carbonate solution, the colour will slowly be changing. And when the end point of the titration has been reached, there will a change in colour. This colour change will be due to the sodium carbonate solution reacting slowly with HCl to form sodium chloride, water and carbon dioxide. When all of the sodium carbonate solution has been used up, then that will indicate the end point of the titration and also the point when the colour completely changes. General Background: An Acid, according the Br�nsted and Lowry theory of acids and bases, is a proton (hydrogen ion, H+) donor. A Base, on the other hand, is defined as a proton (hydrogen ion, H+) acceptor. A good example of a strong acid is sulphuric acid (H2SO4) while sodium hydroxide (NaOH) is an example of a strong base, or alkali - since it is soluble in water. When an acid reacts with a carbonate, it gets neutralized and a salt along with water and carbon dioxide is formed as shown below: Acid + Carbonate ==> Salt + Water + Carbon dioxide The salt produced depends on the acid and carbonate reacting together. ...read more.
Middle
Mean Volume (� 0.10 cm3) 1 0 24.7 24.7 24.8 2 0 24.7 24.7 3 0 24.8 24.8 4 0 24.8 24.8 Mean volume in cm3: (24.7 � 0.10) + (24.7 � 0.10) + (24.8 � 0.10) + (24.8 � 0.10) = (99 � 0.30) � 4 = (24.8 � 0.10) cm3 Mean volume in dm3: = (24.8 � 0.10) � 10-3 dm3 = (0.0248 � 0.0001) dm3 Data processing and presentation: * Chemical equation: 2HCl (aq) + Na2CO3 (aq) ==> 2NaCl (aq) + H2O (l) + CO2 (g) * From the above we can see that 2 moles of HCl react with 1 mole of Na2CO3 to form 2 moles of NaCl, 1 mole of H2O and 1 mole of CO2. * Number of moles of HCl = Given mass (gms) � Molar mass (gms) = 3.65 � (1 + 35.5) = 3.65 � 36.5 = 0.1 mole * Concentration of HCl = No. of Moles (mol) � Volume (dm3) = 0.1 � 1 = 0.1 mol dm-3 * Molar ratio: No. of moles of HCl : No. of moles of Na2CO3 = 2 : 1 (as shown in the chemical equation) Where the no. ...read more.
Conclusion
moles. Percentage Error: 10.1 - 10.0 = 0.1 % Error = 0.1 � 100 = 1 % 10 Evaluation: I must have done this experiment of titration better than the last one, as I obtained a result very close to the actual value with a percentage error of 1 % only. Weaknesses in my method (Sources of error): * There were not that many, but one for sure was that I didn't get a good clamp stand as all the 'good' ones were already taken. I got an old wooden clamp stand which didn't even hold the burette steady (i.e. I couldn't clamp it tight enough). And the clamp stand was also not upright, but leaning a bit forward. This caused the burette to lean forward also, thus causing some errors in reading the volume. Prevention of those errors: * The error of the clamp stand could simply be solved by using a better camp stand, with a good 'clamp' and a strong, upright stand. Further Research: * Some further research could be done with other carbonates such as calcium carbonate (CaCO3) or magnesium carbonate (MgCO3) to find if the number of moles of the water of crystallization is the same in all carbonates ?? ?? ?? ?? Teacher: Minati Sahu 1 Kelwin Joanes - I.B ...read more.
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