Lab 16 B. Buffered Solutions: Solution 6 30.0 mL 0.10 M HOAc +40.0 mL 0.10 M NaO
ID: 1037516 • Letter: L
Question
Lab 16 B. Buffered Solutions: Solution 6 30.0 mL 0.10 M HOAc +40.0 mL 0.10 M NaOAc Data: d yellou 1. Indicator Color 2. PH (Indicator) 3. pH (pH meter)o 5 pH based on color Calculation: 1.76x r 4. pH (calculated ube aucrvled vow of ha to f pH Solution 7-20.0 mL buffer + 2.0 mL 0.100 M HCI Data: 1. Indicator Color e e 2. pH (Indicator) 3. pH (pH meter) 3.5 440 Calculation 4. pH (calculated) Solution 8 -20.0 mL buffer + 2.0 mL 0.10 M NaOH Data light Yalou 1. Indicator Color 2. pH (Indicator) 3. pH (pH meter) Calculation 4. pH (calculated)Explanation / Answer
1. To calculate pH of the solution 6, we use Handerson Hasselbach Equation.
pH = pKa + log [A-]/[HA]
pKa of HOAc = 4.75
[A-] = concentration of NaOAc in soluution =>
concentration of NaOAc used = 0.10 M
volume of NaOAc used = 40.0 ml
Toal volume of solution = 40.0+30.0 = 70.0 ml
[NaOAc] = 0.10*40.0/70.0 = 0.057 M
Similarly for,
[HA] = concentration of HOAc in soluution =>
concentration of HOAc used = 0.10 M
volume of HOAc used = 30.0 ml
Toal volume of solution = 40.0+30.0 = 70.0 ml
[HOAc] = 0.10*30.0/70.0 = 0.043 M
In Handerson Hasselbach equation,
pH = pKa + log [A-]/[HA]
= 4.75 + log (0.057/0.043)
= 4.87
2. To calculate pH of the solution 7,
20.0 ml buffer and 2.0 ml 0.100 M HCl
In 20.0 ml of buffer,
moles of HOAc = concentration*volume = 0.043*0.020 = 8.6*10-4 mol
moles of NaOAc = concentration*volume = 0.057*0.020 = 11.4*10-4 mol
Moles of HCl = concentration*volume = 0.100*0.002 = 2*10-4 mol
Out of HOAc and NaOAC, HCl will react with NaOAc,
so,
OAc- + H+ ------> HOAc
ICE table:
Applying Handerson Hasselbach equation,
pH = 4.75 + log (9.4*10-4)/(10.6*10-4)
= 4.70
3. To calculate pH of the solution 8,
20.0 ml buffer and 2.0 ml 0.100 M NaOH
In 20.0 ml of buffer,
moles of HOAc = concentration*volume = 0.043*0.020 = 8.6*10-4 mol
moles of NaOAc = concentration*volume = 0.057*0.020 = 11.4*10-4 mol
Moles of NaOH = concentration*volume = 0.100*0.002 = 2*10-4 mol
Out of HOAc and NaOAC, NaOH will react with HOAc (because HOAc is an acid and NaOH is a base)
so,
HOAc + OH- ------> H2O + OAc-
ICE table:
Applying Handerson Hasselbach equation,
pH = 4.75 + log (13.4*10-4)/(6.6*10-4)
= 5.06
4. On addition of water, there is no change in pH of the buffer, so pH will remain same.
pH = 4.87 (as shown in part 1)
[HOAc] [H+] [OAc-] Initial 8.6*10-4 11.4*10-4 Change +2*10-4 -2*10-4 Equilibrium 10.6*10-4 9.4*10-4Related Questions
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