Question

Questions: 1. Write the chemical reaction that occurs during the titration. NaOH + HCL NaCl + H₂O 2. Locate in the experiment the definition of equivalence point. Write it here. 3. Read the Plotting the Results section in the experiment and locate the equivalence point on the curve you've drawn using the method described. Mark it on your graph. 4. Using the titration curve you've just drawn determine how many mL of base were required to reach the equivalence point? 29. 15 ML is required to read the equivalence point. 5. How many moles of base does this many mL correspond to? Show calculation. man 1095 moles NaOH molarity n=0.035 0.115 mol NaOH of Naon =0.11SM n=mxv 6. How many moles of acid are contained in your 25.00 mL sample? 7. What is the concentration of your sample in M? Show calculation.

If you could help answer the blanks and check what I have using the graph and info sheet that would be great.

Answer 1

2. Equivalence point is the point where the number of moles of base(NaOH) added becomes equal to the number of moles of acid (HCl) which were present in original unknown concentration HCl solution.

In the given experiment pH is observed at each addition of base. NaOH is a strong base and HCl is a strong acid. In a titration of strong acid versus strong base, at the equivalence point, mixture becomes neutral and pH becomes 7.

Therefore, the volume of base added at pH 7 is called as equivalence point.

In the experiment, volume of NaOH added at equivalence point= 29.75 ml

5. Number of moles of NaOH added to reach equivalence point= Molarity X Volume = 0.115 mol/L X 0.02975 L = 0.00342125 moles

6. At equivalence point, [base] = [acid]

So, number of moles of acid = 0.00342125 moles in 25 ml of the sample

7. Concentration in Molarity = number of moles / volume in litres

= 0.00342125 / 0.025 = 0.13685 M

3. On the curve, volume where pH becomes 7 is the equivalence point. Here in this case at 29.75 ml of base added