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

Lab Report - What volume of concentrated 12N HCl is required to prepare 250 cm3 of 5N HCl solution?

Extracts from this document...

Introduction

Chemistry What volume of concentrated 12N HCl is required to prepare 250 cm3 of 5N HCl solution? Lab Report Experimental Design Aspect 1 Variables Responding Variables - 1. Normality of HCl solution - 12N Controlled Variables - 1. Volume of 5N HCl solution to be prepared - 250 cm3 2. Volume of 12N solution required - 104.16 cm3 Aspect 2 Hypothesis M1*V1 = M2*V2 So, V1 = (M2*V2) / M1 V1 = (5 * 250) / 12 Thus, V1 = 104.16 cm3 Therefore, 104.16 cm3 of 12N HCl is required to prepare 250 cm3 of 5N HCl solution. Aspect 3 Controlling Variables 1. Use a parallax card while measuring the amount of concentrated HCl in order to avoid any sort of error. ...read more.

Middle

Accurately measure out 104.2cm3 of 12N concentrated HCl in a measuring cylinder. 2. Take approximately around 100-120 cm3 of distilled water in the 500 cc beaker and add 104.2 cm3 of Concentrated 12N HCl carefully. Glass rod must be used for stirring. Gloves must be used along with safety goggles as a step of precaution. 3. Cool the beaker when it gets hot under the tap water from time to time and transfer the solution to the 250 cm3 measuring flask. 4. Add the remaining amount of distilled water using the wash bottle to get accurate results. 5. When the water level of the measuring flask reaches the 250 cm3 mark, shake the solution well so that the solution becomes uniform. ...read more.

Conclusion

5N HCl solution to be prepared 250 Conclusion and Evaluation Aspect 1 Conclusion Therefore, in order to prepare 5N HCl we require 104.16 cm3 of 12N HCl. Aspect 2 Evaluating Procedures * Percentage errors- Percentage error= Uncertainty*100/True Value * Volume of 12N concentrated HCl solution = 00.05 * 100 (Measuring flask) 104.5 = 0.049% * Volume of 5N HCl solution made = 00.05 * 100 (Measuring flask) 250 = 0.02% ? Total uncertainty = 0.069% Aspect 3 Improving the Investigation * All the utensils should be washed properly to remove the uncertainty. * Every time use a fresh sample of the chemicals. * Use tested equipments with the least amount of uncertainty. * Make sure you do the experiment carefully to avoid the manual errors. ...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. acid base lab report

    moles of HCl initial - moles of HCl reacted (aka mols NaOH) Mol HCl final = 0.0050 mol HCl - (0.1M)(z) The total volume: Total Volume = initial volume of HCl + volume of NaOH added Total Volume = 0.050 L + z The concentration of HCl: [H+] = mol H+ / total volume 1.6 10-6 = (0.0050 - 0.1z)

  2. Enthalpy of Combustion Lab Report

    Hence the percentage error for this experimental procedure is: = [(1371 - 948) ? 1371] ? 100 = 30.9? This error is large and cannot be accounted for by the uncertainties in the experimental measurements. The major uncertainties lie in the underlying assumptions used in the experimental procedure such as the following: 1.

  1. Aspirin Lab Report

    1 0.868 27.51 44.71 10 Trial 2 0.850 1.20 16.52 10 Trial 3 0.858 16.52 34.60 10 Qualitative information Aspirin 4 Trial 1 0.331 0.31 16.61 10 Trial 2 0.345 16.54 32.11 10 Trial 3 0.292 32.10 47.89 10 Qualitative information Data Processing Titration Cardioton Trial 1 Mass of Aspirin:0.868

  2. Indicator Lab Report - investigating acid-base reactions

    In all experiments involving a weak acid and/or base, there are at least two reactions. The first reaction shows the dissociation of the molecule through a reversible reaction, and the second shows the appropriate ion reacting with its complementary ion to form water.

  1. IB chemistry revision notes

    happens to be fairly close then electrons in it will be induced to move away. A temporary dipole then induces another temporary dipole. These weak forces are the Van der Waal's forces * H * |||||| H --> H * H * F2 Gas Both m.p.t and b.p.t increase Cl2

  2. Change of Potential Difference in Voltaic Cells Lab Report

    To add the remaining water use the water wash bottle. In this way, the 1M CuSO4 solution is ready to use. Throughout this experiment, 5 different concentrations of Copper Sulfate pentahydrate should be prepared to measure the change in voltage.

  1. Titration of Na2CO3.xH2O with HCl

    Thus, a change of colour would indicate the end-point of the reaction. Variables: * Independent variable: There was no independent variable present. * Dependent variable: The dependent variable of this experiment was the volume of dilute acid (HCl) used to neutralize the sodium carbonate solution (Na2CO3).

  2. The chemistry of atmospheric and water pollution.

    The huge influx of industrial waste from factories and sewage from homes greatly increase the concentrations phosphates and nitrate ions as well as heavy metal ions such as cadmium, lead and mercury. Because the level of organic matter has increased, the BOD also increases thus meaning depleting levels of dissolved oxygen available for aquatic organisms.

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