Chemistry 104
posted by Radha on .
The pH of 0.3 HSO3 is:
(How can this problem be solved?)
The final answer is 4.41
However, I keep getting 3.76

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............0.3Mx
Plug in values to Ka equation:
Ka=[x][x]/[0.3Mx]
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............0.3Mx
Plug in values to Ka equation:
Ka=[x][x]/[0.03Mx]
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............0.3Mx
Plug in values to Ka equation:
Ka=[x][x]/[0.3Mx]
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............0.03Mx
Plug in values to Ka equation:
Ka=[x][x]/[0.03Mx]
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.