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Chemistry, 28.02.2020 01:20 kasier4600

Create a graph of pH vs. volume of added NaOH (mL) to the HCl solution of unknown concentration, to produce a titration curve. 2. Find the equivalence point on the graph. What is the pH and equivalence volume of NaOH at this point? 3. a. At the equivalence point, how many moles of 1 M NaOH were added to the flask. To get the number of moles, multiply the molarity by the number of LITERS of NaOH added. b. At the equivalence point, the number of moles of NaOH equals the number of moles of HCl in the flask. How many moles of HCl were in the flask? 4. Using the number of moles of HCl in the flask and the fact that 25 mL of acid were added, calculate the molarity of the HCl solution. (molarity = moles / LITER). 5. An alternative way of calculating the molarity of the HCl solution is to use the fact that as a strong acid, all of the H+ and Cl- dissociate. The pH = -log10([H+]) and thus the concentration of HCl is given by [HCl] = 10^-pH. Using this relationship and the pH of the solution before any NaOH was added, calculate the molarity of the HCl solution. 6. How do the two molarities compare? What could explain any discrepancies? Which calculated molarity is more accurate? Experiment 2 1. Create a graph of pH s. volume of added NaOH (mL) to the acetic acid solution of unknown concentration, to produce a titration curve. 2. Find the equivalence point on the graph. What is the equivalence volume of NaOH at this point? 3. a. Calculate the unknown molarity of the diluted acetic acid from the volumes of acid and base at the equivalence point and the molarity of the NaOH Ma Γ— Va = Mb Γ— Vb. b. Once you find the molarity of your diluted solution use that to calculate the molarity of the original solution using the equation M1 Γ— V1 = M2 Γ— V2 a second time. 4. In experiment 1, you were able to calculate the concentration of the HCl solution using the initial pH. Would this same approach work with the acetic acid? Why or why not? 5. How did the titrations of the two acids compare? Did the results align with the differences between strong and weak acids? Lab NOTES: Ex 1 starting ph= 0.95 start volume 50ml volume46.82 volume dispenced=3.18 ph=12.15 volume 47.15 volume dipenced =2.85ml ph=11.25 a little before but not yet at titration ph 7 47.20 volume voulme dispenced=2.80ml final titration ph 11.25 volume 47.15 volumed dispenced= 2.85 Ex 2 startion ph 2.78 starting volume 50 48,67 volume 1.33 dispenced 47.69 volume 2.31 dispenced 45.50 volume 4.50 dispenced final ph 12.23 volume 46.00 dispenced 4.00ml ph 8.95 right before titrated ph 6.65 46.05 volume 3.95 dispenced final titration ph 8.79 46.00 volume 4.00 dispenced

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