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Preparation and standardisation of HCL & NaOH solutions

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PREPARATION AND STANDARDAIZATION OF HCL & NaOH SOLUTIONS LAB PARTNER: Ryan McPortland October 27, 2008 Introduction: Purpose of this lab was to prepare solutions of acid base and then standardize them by using the titration method. Then use them to determine the unknown concentration of an acid solution. Titration is a procedure that is used to study the stoichiometry of a reaction. Before we start the lab we have to some pre-lab exercises. The first exercise is to calculate the number of grams of NaOH required to make 250 mL of a .1 M solution of NaOH. When we calculate this we get 1g. Then the second exercise was to calculate the number of milliliters of 1.20 M HCL required to make 100.0 mL of a .100 M solutions of HCL. ...read more.


mol .0102 mol .00109 mol Moles HCL .0019 .0102 mol .00109 mol Molarity of HCL .093 M .093 M .092 M Average Morality .093 M - - Concentration of Unknown HCL Solution..........Unknown 2 Sample 1 Sample 2 Sample 3 Volume of HCL 12 mL 12 mL 12 mL Initial Volume NaOH, Vi (mL) Final Volume NaOH used (mL) 16 mL 21.2 mL 26.6 mL Total Volume NaOH, used (mL) 15 mL 5.2 mL 5.4 mL Molarity NaOH .093 M .093 M .093 M Moles NaOH .000484 mol .000484 mol .000502 mol Molarity of HCL .000484 mol .000484 mol .00502 mol Molarity of HCL .040 M .040 M .042 AVERAGE MOLARITY .041 M Conclusion- The titration methods were use to prepare solutions of acid and base and also to determine the unknown concentration of an acid solution. ...read more.


3. Would the calculated molarity of the HCL solutions be higher or lower or not affected if each of the following occurred. Explain your answers A. The buret containing NaOH was rinsed with distilled water but not rinsed with NaOH before being filled. Reduce NaOH concentration, more NaOH would be used. Therefore molarity of HCL would be more. B. The calculated molarity was too high The moles of NaOH used would be high. Therefore HCL would be high C. The tip of the NaOH buret contained an air bubble at the beginning of the titration but not at the end. It would seem like more NaOH was used Therefore HCL higher. 4. Suppose you used the base Ca(OH)2 instead of NaOH. What changes would need be made to the calculations of the HCL concentrations? It would have to double because twice as more OH ions are used. ?? ?? ?? ?? ...read more.

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