Saturday

February 28, 2015

February 28, 2015

Posted by **Radha** on Sunday, May 11, 2014 at 8:02pm.

(How can this problem be solved?)

The final answer is 4.41

However, I keep getting 3.76

- Chemistry 104 -
**Devron**, Sunday, May 11, 2014 at 9:35pmKa 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.

- Chemistry 104 -
**Devron**, Sunday, May 11, 2014 at 9:37pmKa 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.

**Answer this Question**

**Related Questions**

math - for the problem asking to integrate (x^2 - 5x+ 14 )/(x-1) * (x^2+9) I ...

chemistry - What are the equilibrium concentrations of H+,OH- ,(HSO3)-, (SO3)2- ...

Derivatives - If g(x) = tan (5x2), then g ′(x) = ? I've worked this ...

Algebra/ A.S.A.P - Need help. I want to cut an extremely long problem that I ...

Chemistry - Calculate the concentrations of all species in a 0.810 M Na2SO3 (...

chemistry - SF4 ---> SF2 + F2 rate constant (k) = .011 l/molxS [SF4]2 How ...

chemistry...Please Help - I posted this below and received help but im still ...

Chemistry - Water will react most completely as an acid with A. SO3^-2 B. H2BO3...

Chemistry - Calculate the concentration (% W/V) of NaCl solution that was made ...

Physics - Suppose a nucleus of 93X236 fissions into two fragments whose mass ...