When only one concentration at equilibrium is known the concentrations of the others can be calculated using stoichiometry. This is possible because all relationships in a chemical reaction are direct proportions. Calculate the the missing concentrations, then calculate the K, when a 0.050 M solution of CH3NH₂ is made, and at equilibrium [CH3NH₂] = 0.040 M. CH3NH₂ + H₂O CH3NH3 + OH-¹ 0 0 start: 0.050 M equil: 0.045 M 0.005 M The amount of CH3NH₂ that was converted into product was 0.050 M -0.045 M = For each 1 mol CH3NH₂+1 (0.005 M CH3NH2 )(---------------) = 1 mol CH3NH₂ 1 mol OH-1 (0.005 M CH3NH2)(------- Keq = 1 mol CH3NH₂ of CH3NH₂ converted into products, there will be 0.005 M of CH3NH3*1 and [CH3NH3*1¹][OH-¹] [ -) = ][ CH3NH3*¹ OH-1 [CH3NH₂] [ (notice that the information to calculate both the pOH and pK, are provided) E. CH3NH₂ F. CH3NH₂+¹ A. HF B. H₂O C. F-1 D. H30*1 J. 0.005 M K. 0.040 M L. 0.045 M M. 0.00056 M N. 0.0056 M G. OH-1 H. 0.030 M. O. 0.11 M P. 0.50 M 1. 0.050 M of OH1, or
When only one concentration at equilibrium is known the concentrations of the others can be calculated using stoichiometry. This is possible because all relationships in a chemical reaction are direct proportions. Calculate the the missing concentrations, then calculate the K, when a 0.050 M solution of CH3NH₂ is made, and at equilibrium [CH3NH₂] = 0.040 M. CH3NH₂ + H₂O CH3NH3 + OH-¹ 0 0 start: 0.050 M equil: 0.045 M 0.005 M The amount of CH3NH₂ that was converted into product was 0.050 M -0.045 M = For each 1 mol CH3NH₂+1 (0.005 M CH3NH2 )(---------------) = 1 mol CH3NH₂ 1 mol OH-1 (0.005 M CH3NH2)(------- Keq = 1 mol CH3NH₂ of CH3NH₂ converted into products, there will be 0.005 M of CH3NH3*1 and [CH3NH3*1¹][OH-¹] [ -) = ][ CH3NH3*¹ OH-1 [CH3NH₂] [ (notice that the information to calculate both the pOH and pK, are provided) E. CH3NH₂ F. CH3NH₂+¹ A. HF B. H₂O C. F-1 D. H30*1 J. 0.005 M K. 0.040 M L. 0.045 M M. 0.00056 M N. 0.0056 M G. OH-1 H. 0.030 M. O. 0.11 M P. 0.50 M 1. 0.050 M of OH1, or
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Question
![When only one concentration at equilibrium is known the concentrations of the others can be calculated using stoichiometry. This is possible because all relationships in a chemical reaction are
direct proportions. Calculate the the missing concentrations, then calculate the K, when a 0.050 M solution of CH3NH₂ is made, and at equilibrium [CH3NH₂] = 0.040 M.
CH3NH2 + H₂O → CH³NH₂+¹ + OH-¹
0
0
start: 0.050 M
equil: 0.045 M
0.005 M
The amount of CH3NH₂ that was converted into product was 0.050 M -0.045 M =
For each
1 mol CH3NH3+¹
(0.005 M CH3NH₂ X-
Kea
1 mol CH3NH₂
(0.005 M CH3NH2)(-
1 mol OH-1
1 mol CH3NH₂
[CH3NH3 +¹][OH-¹]
[
of CH3NH₂ converted into products, there will be 0.005 M of CH3NH3*¹ and
-) =
][
CH3NH3+1
OH-1
[CH3NH₂]
[
]
(notice that the information to calculate both the pOH and pk, are provided)
A. HF B. H₂O C. F-1
D. H30+1
J. 0.005 M K. 0.040 M L. 0.045 M
E. CH3NH₂ F. CH3NH₂+¹
M. 0.00056 M N. 0.0056 M
G. OH-1 H. 0.030 M I. 0.050 M
O. 0.11 M P. 0.50 M
of OH-1, or](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Feab2d2d3-91b0-42ef-9603-f34b4b53ecf8%2Fbb61d84f-5c4b-4b45-b93e-25d3f9a59035%2Fj4lt70k_processed.png&w=3840&q=75)
Transcribed Image Text:When only one concentration at equilibrium is known the concentrations of the others can be calculated using stoichiometry. This is possible because all relationships in a chemical reaction are
direct proportions. Calculate the the missing concentrations, then calculate the K, when a 0.050 M solution of CH3NH₂ is made, and at equilibrium [CH3NH₂] = 0.040 M.
CH3NH2 + H₂O → CH³NH₂+¹ + OH-¹
0
0
start: 0.050 M
equil: 0.045 M
0.005 M
The amount of CH3NH₂ that was converted into product was 0.050 M -0.045 M =
For each
1 mol CH3NH3+¹
(0.005 M CH3NH₂ X-
Kea
1 mol CH3NH₂
(0.005 M CH3NH2)(-
1 mol OH-1
1 mol CH3NH₂
[CH3NH3 +¹][OH-¹]
[
of CH3NH₂ converted into products, there will be 0.005 M of CH3NH3*¹ and
-) =
][
CH3NH3+1
OH-1
[CH3NH₂]
[
]
(notice that the information to calculate both the pOH and pk, are provided)
A. HF B. H₂O C. F-1
D. H30+1
J. 0.005 M K. 0.040 M L. 0.045 M
E. CH3NH₂ F. CH3NH₂+¹
M. 0.00056 M N. 0.0056 M
G. OH-1 H. 0.030 M I. 0.050 M
O. 0.11 M P. 0.50 M
of OH-1, or
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