Posted by Radha on Sunday, May 11, 2014 at 8:02pm.
Ka of HSO3^-=6.3 x 10^-8
HSO3^- --------> H^+ + SO3^-
Ka=[H^+][SO3^-]/[HSO3^-]
E of the ICE chart is the following:
[H^+]..[SO3^-]...[HSO3^-]
x............x............0.3M-x
Plug in values to Ka equation:
Ka=[x][x]/[0.3M-x]
I am going to assume x is small and ignore, but will have to check the assumption afterwards.
6.3 x 10^-8=x^2/[0.3M]
([0.3M]*Ka)^1/2=x
x=1.37 x 10^-4 M
1.37 x 10^-4 M/0.3M=0.04%
The assumption checks out.
x=H^+
and pH=-log[H^+]
pH=-log[1.37 x 10^-4 M]=3.86
pH=3.86
Now, pay attention and see what happens when I change the concentration from 0.3M to 0.03M:
E of the ICE chart is the following:
[H^+]..[SO3^-]...[HSO3^-]
x............x............0.3M-x
Plug in values to Ka equation:
Ka=[x][x]/[0.03M-x]
I am going to assume x is small and ignore, but will have to check the assumption afterwards.
6.3 x 10^-8=x^2/[0.03M]
([0.03M]*Ka)^1/2=x
x=4.35 x 10^-5 M
4.35 x 10^-5 M/0.03M=0.14%
The assumption checks out.
x=H^+
and pH=-log[H^+]
pH=-log[4.35 x 10^-5 M]=4.36
pH=4.36
I have never seen a HSO3^- concentration that high, which is were the error has occurred. Next time, include the Ka value that is given to you. This seems like a made up problem that your teacher didn't check or you made a typo. I don't think that a 0.3M solution of HSO3^- can be made.
Ka of HSO3^-=6.3 x 10^-8
HSO3^- --------> H^+ + SO3^-
Ka=[H^+][SO3^-]/[HSO3^-]
E of the ICE chart is the following:
[H^+]..[SO3^-]...[HSO3^-]
x............x............0.3M-x
Plug in values to Ka equation:
Ka=[x][x]/[0.3M-x]
I am going to assume x is small and ignore, but will have to check the assumption afterwards.
6.3 x 10^-8=x^2/[0.3M]
([0.3M]*Ka)^1/2=x
x=1.37 x 10^-4 M
(1.37 x 10^-4 M/0.3M)*100=0.04%
The assumption checks out.
x=H^+
and pH=-log[H^+]
pH=-log[1.37 x 10^-4 M]=3.86
pH=3.86
Now, pay attention and see what happens when I change the concentration from 0.3M to 0.03M:
E of the ICE chart is the following:
[H^+]..[SO3^-]...[HSO3^-]
x............x............0.03M-x
Plug in values to Ka equation:
Ka=[x][x]/[0.03M-x]
I am going to assume x is small and ignore, but will have to check the assumption afterwards.
6.3 x 10^-8=x^2/[0.03M]
([0.03M]*Ka)^1/2=x
x=4.35 x 10^-5 M
(4.35 x 10^-5 M/0.03M)*100=0.14%
The assumption checks out.
x=H^+
and pH=-log[H^+]
pH=-log[4.35 x 10^-5 M]=4.36
pH=4.36
I have never seen a HSO3^- concentration that high, which is were the error has occurred. Next time, include the Ka value that is given to you. This seems like a made up problem that your teacher didn't check or you made a typo. I don't think that a 0.3M solution of HSO3^- can be made.