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............0.3M-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.
To solve this problem, you need to understand the concept of pH and how it relates to acidic or basic solutions.
pH is a measure of the acidity or alkalinity of a solution. It is defined as the negative logarithm (base 10) of the concentration of hydrogen ions (H+) in a solution. The pH scale ranges from 0 to 14, where 0 is highly acidic, 7 is neutral, and 14 is highly basic.
In this problem, you are given a concentration of 0.3 HSO3-. To find the pH, you first need to determine the concentration of H+ ions in the solution.
HSO3- is the conjugate base of the weak acid H2SO3. H2SO3 is a diprotic acid, which means it can donate two protons (H+ ions). When H2SO3 loses one proton, it forms HSO3-.
To find the concentration of H+ ions in the solution, you need to consider the equilibrium reaction of HSO3-:
HSO3- + H2O ⇌ H3O+ + SO32-
Using an equilibrium expression, we can write the equation as:
[H3O+] = K × [HSO3-]
K is the equilibrium constant for this reaction. Since HSO3- is a weak acid, its dissociation constant (Ka) can be used as the equilibrium constant for this reaction. The Ka value for HSO3- is 1.2 x 10^-2.
Now, substitute the given concentration [HSO3-] = 0.3 into the equation:
[H3O+] = (1.2 x 10^-2) × (0.3)
[H3O+] = 3.6 x 10^-3
To find the pH, take the negative logarithm (base 10) of [H3O+]:
pH = -log([H3O+]) = -log(3.6 x 10^-3) = 2.44
Therefore, the pH of 0.3 HSO3- is 2.44.
It appears that your calculated value of 3.76 is incorrect. Double-check your calculations or verify the given concentration to ensure accuracy.