Missing in your question Ka2 =6.3x10^-8
From this reaction:
H2SO3 + H2O ↔ H3O+ + HSO3-
by using the ICE table :
H2SO3 ↔ H3O + HSO3-
intial 0.6 0 0
change -X +X +X
Equ (0.6-X) X X
when Ka1 = [H3O+][HSO3-]/[H2SO3]
So by substitution:
1.5X10^-2 = (X*X) / (0.6-X) by solving this equation for X
∴ X = 0.088
∴[H2SO3] = 0.6 - 0.088 = 0.512
[HSO3-] = [H3O+] = 0.088
by using the ICE table 2:
HSO3- ↔ H3O + SO3-
initial 0.088 0.088 0
change -X +X +X
Equ (0.088-X) (0.088+X) X
Ka2= [H3O+] [SO3-] / [HSO3-]
we can assume [HSO3-] = 0.088 as the value of Ka2 is very small
6.3x10^-8 = (0.088+X)*X / 0.088
X^2 +0.088 X - 5.5x10^-9= 0 by solving this equation for X
∴X= 6.3x10^-8
∴[H3O+] = 0.088 + 6.3x10^-8
= 0.088 m ( because X is so small)
∴PH= -㏒[H3O+]
= -㏒ 0.088 = 1.06
Answer:
<h2>36.09 L</h2>
Explanation:
The initial volume can be found by using the formula for Boyle's law which is
where
P1 is the initial pressure
P2 is the final pressure
V1 is the initial volume
V2 is the final volume.
Since we're finding the initial volume
We have
We have the final answer as
<h3>36.09 L</h3>
Hope this helps you
Water containing carbonic acid and calcium
The awnser is D or the 4th one
Answer:
D
Explanation:
The correct thing to do in this case would be to <u>repeat the experiment.</u>
The scientist would need to repeat the experiment in order to double-check the accuracy. If the accuracy is indeed doubtful, he/she can be able to trace the source of the error by repeating the experiment.
The correct option is D.
