V PPost-lab Write a communication-style report describing this experiment (no la
ID: 563812 • Letter: V
Question
V PPost-lab Write a communication-style report describing this experiment (no laundry lists of materials or methods: we all have this protocol!) and incorporating the answers to the following post-lab question in an appendix, after your references. 1. Explain any shifts in pH you observe with the addition of the MgCl2 and NaCl to the 10 mL aliquot of phosphate buffer. Was one salt solution more effective than the other, why? Please state the net ionic equations to help fortify your answer on their effectiveness. 2. Answer these questions concerning your Tris Buffer a. How much HCI solution was used to adjust the pH of the Tris buffer? How did that b. If you were unable to titrate the Tris solution to pH 8, what was the actual pH of 3. What was the actual pH of your group's phosphate buffer? How did it compare to the compare to your calculated amount? your Tris buffer? tabulated value? 4. What would the pH be of a solution obtained by mixing 0.2 ml 1 M HCI with 10 ml distilled water? What would the pH be of a solution obtained by mixing 0.2 ml 1M HCI with 10 ml of 10 nM NaOH(aq)? How do these values compare to the pH values you obtained when you added acid to your buffers? 5. Talk with your classmates in other groups: which phosphate buffer's pH changed the least when acid was added? Which phosphate buffer's pH did you expect to change the least? 6. When your group added imidazole to your tris buffer, what happened to the pH? Did imidazole act as an acid or a base? When your group added imidazole to your phosphate buffer, what happened to the pH? Did imidazole act as an acid or a base? Did imidazole act the same way for each group?Explanation / Answer
4.
a) you have 0.2ml, equals to 0.0002 liters of solution of 1 M
calculate thenumber of moles with
moles = Molarity * volume
moles = 1 * 0.0002 = 0.0002 moles of HCl
the new volume will be 10 ml + 0.2 ml = 10.2 ml so
this is 10.2 / 1000 = 0.0102 liters
new molarity is moles/ volume
0.0002 moles of HCl / 0.0102 liters = 0.0196 M
PH = -log (0.0196) = 1.7075
part b
0.2 ml of 1 M HCl has 0.0002 moles
10 ml (0.01 liters) of 10 nM ( 1 x 10-8M) of NaOH
calculate moles
moles = 0.01 * 1 x 10-8M = 1 x 10-10 moles of NaOH
get the difference
0.0002 - 1 x 10-10 , this is approximately 0.0002 because the moles of NaOH able to neutralize the HCl is very small
new volume = 10 + 0.2 ml = 10.2 ml = 0.0102 liters
Molariy of HCl = 0.0002 / 0.0102 = 0.0196 M
ph = -log (0.0196) = 1.7075
the rest is about the ph of a buffer from a previous experiment
*hope this helps =)
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