Suggested Data Table
(Partially completed)
1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | |
Vol HX Init., mL |
Vol
OH– added for this increment, mL |
Total Vol OH–, mL |
Total
Vol Soln, mL |
[HX]i | [X–]eq | [HX]eq | [X–]/[HX] | log | Meas'd pH | Calc'd pKa | |
0 | 50.0 | 0.0 | 0.0 | 50.0 | --- | --- | --- | --- | --- | --- | --- |
1 | 50.0 | 10.0 | 10.0 | 60.0 | |||||||
2 | 50.0 | 5.0 | 15.0 | 65.0 | |||||||
3 | 50.0 | 5.0 | 20.0 | 70.0 | |||||||
4 | 50.0 | 10.0 | |||||||||
5 | 50.0 | ||||||||||
6 | 50.0 | ||||||||||
please help
Solution 1.
pH = 2.15
i.e. -Log[H+] = 2.15, i.e. [H+] = 10-2.15 = 7.08*10-3 M
i.e. [X-]eq = 7.08*10-3 M
Now, [HX]eq = 0.1 M - 7.08*10-3 M = 0.093 M
Now, Ka = [H+][X-]eq/[HX]eq = (7.08*10-3)*(7.08*10-3)/0.093 = 5.394*10-4
Now, pKa = -Log(Ka) = -Log(5.394*10-4) = 3.268
Log[X-eq/HXeq] = Log(7.08*10-3/0.093) = -1.118
Suggested Data Table (Partially completed) 1 2 3 4 5 6 7 8 9 10 11...
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