• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

The purpose of this experiment is to find the composition of a sample of sodium carbonate mixture by titration.

Extracts from this document...


Name : CHAN CHI HEI Class: 6LS Class no :1 Title: Double titration Date of experiment : Aim The purpose of this experiment is to find the composition of a sample of sodium carbonate mixture by titration. Theory Double indicator titration is used in checking the composition of a mixture of sodium carbonate and sodium hydrogen carbonate. First, no of mol of Na2CO3 in mixture can be directly calculated in the data of first titration. For the calculation of no of mol of NaHCO3 need to involve the data from both 1st and 2nd data , because the first titration will give out NaHCO3 also and it will contaminated with the NaHCO3 in the mixture. Requirements watch glass weighing bottle spatula sodium carbonate Na2CO3 250 cm3 beaker, wash bottle of distilled water glass rod 250 cm3 volumetric flask filter funnel, dropper Procedure 1. 10.00 g of the sodium carbonate is transferred onto the watch glass and weighted in the nearest 0.01g 2. A standard solution is prepared. 3. Burette is rinsed with HCL and then filled with the acid. ...read more.


used(I)/cm3 7.28 Methyl Orange Trial 1st 2nd 3rd Final Burette Reading(II)/cm3 49.4 34.0 34.2 34.0 Initial Burette Reading(II)/cm3 30.1 12.5 12.5 12.05 Volume used(II)/cm3 19.3 21.5 21.7 21.95 Mean Volume of HCl(aq) used(II)/cm3 21.7 First stage of titration Na2CO3(aq) + HCl(aq) >> NaHCO3(aq) + NaCl(aq) Since HCL : Na2CO3 = 1:1 No of mol of Na2CO3 = Mean volume 1 of HCL used x Concentration of HCL = 7.28 /1000 x 0.5 = 3.64 x 10-3 mol Second stage of titration NaHCO3(aq) + HCl(aq) >> NaCl(aq) + H2O(l) + CO2(g) No of mole of NaHCO3 = Mean volume 2 of HCL used x Concentration of HCL = 21.7/1000 x 0.5 =0.01085 mol Calculation From above , No of mole of Na2CO3 in 250 cm3 solution = 3.64 x 10-3 mol No of mole of NaHCO3 in the 250cm3 solution = (No of mole of NaHCO3 in second stage) - (No of mole of NaHCO3 in first stage) = 0.01085 - 3.64 x 10-3 = 7.21 x 10-3 mol Mass of Na2CO3 = 0.0364 x 106 = 3.8584 g Which is around 3.9 g Mass of NaHCO3 = 0.0721 x 84 = 6.0564 g ...read more.


After that , methyl orange is used because NaHCO3 is a weak base . It changes from red below pH 3.1 to yellow above pH 4.4. Therefore , in the second part of titration, when the indicator turn from yellow to orange, the solution is in acidic but not exactly neutral . Therefore , the result obtained from above is not accurate. If we want to get a more accurate result , ph meter should be used but not indicator. The calculation above has a little bit complicated. Since Na2CO3 only involve in the first titration , no of mol in Na2CO3 will exactly equal to the no of mol of HCL used. After that , we need to calculate No of mole of NaHCO3 in the mixture by using the (No of mole of NaHCO3 in second stage) - (No of mole of NaHCO3 in first stage) because the No of mol of NaHCO3 in second stage contain both NaHCO3 from the mixture and that from the first titration. Thus , we need to minus the no of mole of NaHCO3 in first stage to obtain the no of mol of NaHCO3 in the mixture. ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our International Baccalaureate Chemistry section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related International Baccalaureate Chemistry essays

  1. Calcium Carbonate and Hydrochloric Acid

    a too late reading, meaning that the recording took place a second or two later than it should have. The same analysis (as the ones for fig.3) can be applied to fig.4 and the two anomalies on that graph. The 0,0149g reading (on the graph)

  2. To calculate the percentage composition of the mixture of Na2CO3 + ...

    Now, to calculate the mass of NaHCO3 can simply be calculated by unitary method as follows - Let the mass of NaHCO3 be X 168 62 X 0.222 Therefore X = 0.602 + 0.001 g As we know that the mass of the mixture before heating was 0.837 + 0.002g

  1. Aim. To find the identity of X(OH)2 (a group II metal hydroxide) by determining ...

    The percentage error of Ca(OH)2 = [(0.170 - 0.145)/0.170] X 100 = (0.025/0.170) X 100 = 14.705% Throughout the experiment there were systematic errors and random errors that were met. Uncertainties/limitations Error Type of error Quantity of error Explanation for error Improvements Measurement in burette Systematic error +/- 0.05cm3 Equipment

  2. Preparation of ear drops practical. Titration of NaHCO3

    2. Place this in a conical flask. 3. Add 2-3 drops of methyl orange to be an indicator 4. Then titrate eardrops against 0.1M from the burette.

  1. Determination of potassium hydrogen carbonate into potassium carbonate

    + 1.01 = 100.11 � 4 % Moles = 3.555/ 100.11 = 0.0354 � 3 % The enthalpy change is required to be calculated. Therefore the enthalpy change can be calculated by the following equation: Energy change = X

  2. To determine the molecular mass of an unknown alkali metal carbonate, X2CO3.

    Some other possible shortcomings to this experiment are also listed below: 1. The solid substance Z might not be pure Sodium Carbonate and some impurities present could have led to the overstatement of the value for the molecular mass of X2CO3.

  1. Biodiesel Investigation - How the concentration of Potassium Hydroxide solution would affect the yield ...

    Furthermore, all glassware should be kept away from the edges of tables to prevent risk of damage. KOH ? methanol mixture is labeled as an irritant and mildly Independent Variable: Concentration of KOH ? Methanol Mixture as % value. Fixed Variables: * Temperature of Oil * Volume of Oil Used

  2. Chemistry Titration Acid Base Lab

    For instance, when base was being added to the vinegar solution with the indicator Bromocresol green, the color of the solution turned from yellow to green. When the green color is seen, the end point has been reached however, if the color becomes blue then over-titration has taken place, therefore shifting equilibrium even further to the right.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work