Chemistry 104

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

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asked by Radha
  1. 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.

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    posted by Devron
  2. 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.

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    posted by Devron

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