The measured HCl was transferred to a 250mL Erlenmeyer flask 5 Three drops of

The measured hcl was transferred to a 250ml

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The measured HCl was transferred to a 250mL Erlenmeyer flask 5. Three drops of the indicator were added to the flask 6. The magnetic stirrer was added to the solution in the flask 7. The magnetic stir plate was placed under the buret 8. The beaker filled with the 16 mL of the HCl solution was placed on the magnetic stir plate 9. The stir plate was turned on to the second setting 10. The stopcock (buret knob) was turned until the flow of the NaOH was dripping at a steady rate 11. When the solution began to turn pink, the stop cock was closed to stop the flow of NaOH 12. If the solution stayed pink for at least 30 seconds, the final buret reading was recorded; If not, additional drop(s) were added to achieve a light pink tone for 30 seconds 13. Steps 1-13 were repeated for each trial, with different amounts of HCL used per trial Step 2: Finding the Molarity of Each Trial 1. The volume of the NaOH was multiplied to the molarity of the NaOH to solve for the moles of NaOH used 2. The moles of NaOH were then converted into moles of HCl using stoichiometry and the chemical formula 3. The molarity of the HCl solution was found through the molarity formula; by dividing the computed amount of moles of HCl by the amount of HCl (titrand) used 4. Steps 1-3 were repeated for all the trials Step 3: The Average Molarity 1. All the molarity values were added together 2. The value found in step one was divided by 3.00 to get the average molarity of the HCl DATA AND OBSERVATIONS: Trial Initial Buret Reading (ml) Final Buret Reading (ml) Amount of titrant used (mL) (NaOH) Amount of titrand used (mL) (HCl) Concentration of titrand (HCl) 1 0.2 17.5 17.3 16 .234 M 2 0.5 11.5 11.0 10 .238 M 3 0.4 22.2 21.8 20 .235 M Average Concentration: .236 M
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CALCULATIONS: NaOH + HCl → NaCl + H2O Note: the concentration of the NaOH solution used was 0.216 M Trial 1: 0.216 M = → 0.00374 mol NaOH mol NaOH 0.0173 L NaOH ( )( ) = 0.00374 mol HCl 1 0.00374 mol NaOH 1 mol HCl 1 mol NaOH = .234 M 0.016 L HCl 0.00374 mol HCl Trial 2: 0.216 M = → 0.00238 mol NaOH mol NaOH 0.0110 L NaOH ( )( ) = 0.00238 mol HCl 1 0.00238 mol NaOH 1 mol HCl 1 mol NaOH = .238 M 0.010 L HCl 0.00238 mol HCl 0.216 M = → 0.0047088 mol NaOH mol NaOH 0.0218 L NaOH ( )( ) = 0.0047088 mol HCl 1 0.0047088 mol NaOH 1 mol HCl 1 mol NaOH = .235 M 0.020 L HCl 0.0047088 mol HCl ANALYSIS AND DISCUSSION: 1.When you titrate an acid with a base, you are advised to titrate to the faintest pink color. Why is this done?
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  • Winter '20
  • pH, mol HCl

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