Part B Using the method of successive approximations, what is [H30 + ] in 0.500
ID: 958455 • Letter: P
Question
Part B Using the method of successive approximations, what is [H30 + ] in 0.500 M HCIO 2 (Ka-1.1 x 10-2 ) and how many iterations (n) are needed to validate that a constant value had been obtained to two significant figures? Express [H3 O Express [H0'] using twa using two significant figures, followed by a comma and the number of iterations as an integer. [H301, n = Submit Hints My Answers Give Up Review Part Incorrect; Try Again The z values in iterations 2 and 3 do not match. Therefore you need more than three iterations. Part C Using the method of successive approximations, what is |H,0 in 0.00500 M CH CICOOH(a)(14x 10-) and how many iterations (n) are needed to validate that a constant value had been obtained to two significant figures? ..imati rst what ishadbe e ind tained tot orignic ooHraq) (K.-1.4× 10-3) and how many Express (H3O"] using two significant figures, followed by a comma and the number of iterations as an integer. [H3O+], n = SubmitHints My Answers Give Up Review Part ContinueExplanation / Answer
Part B
I suggest you to use a Excel file:
iteration no.
[HA], not dissociated
[H+]=( Ka x [HA])^0.5
corrected[HA]=0.500 - [H+]
1
0.5000
0.074
0.4258
2
0.4258
0.068
0.4316
3
0.4316
0.069
0.4311
4
0.4311
0.069
0.4311
5
0.4311
0.069
0.4311
6
0.4311
0.069
0.4311
[H3O+] = 0.069 M , after 4 iterations is confirmed unchanged
Part C
iteration no.
[HA], not dissociated
[H+]=( Ka x [HA])^0.5
corrected[HA]=0.00500 - [H+]
1
0.0050
0.0026
0.0024
2
0.0024
0.0018
0.0032
3
0.0032
0.0021
0.0029
4
0.0029
0.0020
0.0030
5
0.0030
0.0020
0.0030
6
0.0030
0.0020
0.0030
7
0.0030
0.0020
0.0030
8
0.0030
0.0020
0.0030
[H3O+] = 0.020 M , after 5 iterations is confirmed unchanged
iteration no.
[HA], not dissociated
[H+]=( Ka x [HA])^0.5
corrected[HA]=0.500 - [H+]
1
0.5000
0.074
0.4258
2
0.4258
0.068
0.4316
3
0.4316
0.069
0.4311
4
0.4311
0.069
0.4311
5
0.4311
0.069
0.4311
6
0.4311
0.069
0.4311