(2) A student followed the procedure described but ued a 0.980 M solution of a w
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Question
(2) A student followed the procedure described but ued a 0.980 M solution of a weak acid (HAn) and 0.510 m NaOH solution.
our professor is pregnant, so we haven't had anyone teach us....help would be greatly appreciated
A student followed the procedure described in this module but used a 0.980M solution of a weak acid (HAn) and 0.510M NaOH solution. The pipet calibration data were: The number of drops of each of the solutions added to the wells and the corresponding pH readings are given in Table 4.Explanation / Answer
Answer for question (1):
No of drops in 1 ml usually varies from 20 to 60 drops. However it’s widely accepted 1 ml has 20
drops in metric system.
1 drop = 0.05 ml
Eeach drop of Han, NaOH and H2O delivered
in litres = 0.05 ml/1000 = 0.000050 L = 5 x 10-5L = 50 µL
Answer for question (2):
In set A-1 volumes of each of Han, NaOH and H2O delivered = 20 drops
Solution of Han delivered in litres for wells A-1, A-2, A-3 and A-4 is 20 drops in each well.
Since 1 drop = 0.05 ml
Solution of Han delivered in litres for wells A-1, A-2, A-3 and A-4
is 20 drops x 0.05 ml = 1 ml = 0.001 L
Similarly for rest of the solutions the volumes are calculated in Litres and presented them in the following table.
A-1, (L)
A-2 (L)
A-3 (L)
A-4 (L)
HAn solution
0.001
0.001
0.001
0.001
NaOH solution
0.00125
0.001
0.00075
0.0005
H2O added
0.00075
0.001
0.00125
0.0015
Answer for question (3):
The total volume of mixture of all solutions when summed up gives 0.003 L as shown in the following table.
A-1, (L)
A-2 (L)
A-3 (L)
A-4 (L)
HAn solution
0.001
0.001
0.001
0.001
NaOH solution
0.00125
0.001
0.00075
0.0005
H2O added
0.00075
0.001
0.00125
0.0015
Total volume of all solutions (L)
0.003
0.003
0.003
0.003
Answer for question (4):
No of moles = volume in Litres x molarity.
Since molarity of Han solution is 0.98 and when it’s multiplied by the volume added in Litres that will give the number of moles of Han added in each well mixture and are calculated accordingly and presented them in the following table.
A-1
A-2
A-3
A-4
0.980 M Han solution added (L)
0.001
0.001
0.001
0.001
Initial/ number of moles of Han in each well mixture (VxM)
0.00098
0.00098
0.00098
0.00098
Answer for question (5):
Initial number moles of Han is equal to that of number of moles of HAn present in each equilibrium mixture. The answer is same as for the question 4.
Answer for question (6):
The balanced equation for reaction of HAn with NaOH is represented as follows:
HAn (aq) + NaOH (aq) ç===============è AnNa (aq) + H2O (l)
Number of moles of undissociated HAn present in each equilibrium mixture:
= Initial number of moles HAn used – No. of moles of NaOH used in each well.
= 0.00098 - No. of moles of NaOH used in each well
No of moles of NaOH used in each well = volume in Litres x molarity.
Molarity of NaOH added = 0.510 M
Volume of NaOH added in each well is as given in the following table. Using this data the number moles of NaOH used can be obtained which is used in getting the undissociated HAn present in each equilibrium mixture of each well. The data is presented in the following table.
A-1
A-2
A-3
A-4
Initial number of moles of Han in each well
0.0009800
0.0009800
0.0009800
0.0009800
No of moles of NaOH added in each well
0.0006375
0.0005100
0.0003825
0.0002550
No of moles of undissociated Han in each well present
0.0003425
0.0004700
0.0005975
0.0007250
Answer for question (7):
For calculating the equilibrium concentration of HAn we need to divide number of moles of undissociated HAn in each well mixture by total well volume of the mixture of all (HAn, NaOH and H2O) solutions.
Equilibrium concentration of HAn = number of moles of undissociated Han/ total well volume of the mixture in litres
Accordingly the data is presented in the following table
A-1
A-2
A-3
A-4
No of moles of undissociated Han in each well present
0.0003425
0.0004700
0.0005975
0.0007250
Total volume (L) of all solutions in each well
0.003
0.003
0.003
0.003
Molarity (M) of Han at equilibrium
0.1142
0.1566
0.1992
0.2416
Answer for question (8):
Reaction of HAn with NaoH is as follows:
HAn (aq) + NaOH (aq) ç========è H2O (l) + NaAn (aq)
Since 1 mole of NaOH reacts with 1 mole of HAn and gives 1 mole of NaAn, the concentration of An- at equilibrium is same as the concentration of NaOH used at equilibrium which is given in the following table.
A-1
A-2
A-3
A-4
No of moles of NaOH added in each well
0.0006375
0.0005100
0.0003825
0.0002550
Total volume of all solutions
0.003
0.003
0.003
0.003
Molarity of An- at equilibrium
0.2125
0.17
0.1275
0.085
Answer for question (9):
pH = - log [H+]
[H+] = 10-pH
Accordingly the data is presented in the following table.
A-1
A-2
A-3
A-4
pH
4.23
3.93
3.68
3.41
equilibrium concentration of H+ (M)
0.0000588
0.000117
0.000209
0.000389
Answer for question (10):
Ka of a weak acid is calculated using the following formula. The values obtained for each of the species for each of the wells is tabulated and finally Ka for each of the well is calculated.
Ka = [H+][An-]/[HAn]
A-1
A-2
A-3
A-4
equilibrium concentration of H+ (M)
0.0000588
0.000117
0.000209
0.000389
Molarity of An- at equilibrium
0.2125
0.17
0.1275
0.085
Molarity of undissociated HAn at equilibrium
0.1142
0.1566
0.1992
0.2416
Ka of weak acid
1.09 e-4
1.27 e-4
1.34 e-4
1.36 x e-4
Answer for question (11):
Mean Ka for the weak acid = [Ka of A-1 + Ka of A-2 + Ka of A-3+ Ka of A-4]/4
Accordingly
Mean Ka for the weak acid = [0.000109 + 0.000127 + 0.000134 + 0.000136]/4
Mean Ka for the weak acid = 0.0001265 = 1.265 x 10-4
Answer for question (12):
Ka for the unknown weak acid is not included in the table. However from literature the closest Ka for the observed Ka is that of citric acid Ka = 1.38 x 10-4.
So the unknown weak acid is citric acid.
Percent error of Ka calculated = Standard Ka – Observed Ka/Standard Ka x 100
= 0,000138 – 0.0001265 / 0.000138 x 100
Percent error of Ka calculated = 8.33%
A-1, (L)
A-2 (L)
A-3 (L)
A-4 (L)
HAn solution
0.001
0.001
0.001
0.001
NaOH solution
0.00125
0.001
0.00075
0.0005
H2O added
0.00075
0.001
0.00125
0.0015
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