3. Make your first ICE Table (Initial, Change, Equilibrium)
...[H2SO3] <-----> [H3O+] + [HSO3-]
I...0.60M...............0............. (a 0 should be here)
C...-X....................+X........ (a +X should be here... having a hard time showing)
E...0.60-X..............X............(a... X should be here)
We will be making products here, since we started with the reactant, so the products start out with nothing initially and have some number (X) added to each of them. There is only 1X added/subtracted since the mole to mole ratio is 1 to 1.
4. Plug in your equilibrium values into your expression for Ka1 and...
1.5 x10⁻² = (X)(X) / (0.60-X)
...solve for X. This will be a quadratic equation since you cannot use the small X approximation. For my equation, I got:
X² + 1.5x10⁻²X -- 0.009
You can work this through the quadratic formula, or if you have a graphing calc- typically a TI 83 or 84, you can press the MATH button, 0, and it will take you to an equation solver. Enter in your quadratic and press Enter. The next screen has your quadratic on the first line, an X= on the second line, and "bound= ..." on the last line. We only need the second line on this screen. For X= you will need to enter in .001 (or more zeros- I always enter in a handful just in case.
Once you have entered in your number in for X= , hit "ALPHA" and Enter (which also means "solve").
It will spit out some decimal for you. DON'T forget to scroll ALL the way to the right on this line, where there is often an E-- #.
For my X, I got 0.088. You can now plug this back into your ICE table to get concentrations of all species present. You really only need the concentrations for H3O+ and HSO3- for this problem though.
5. Write out your second chemical equation and expression for Ka2. Since this is a diprotic acid, and we want to get all the H atoms off the acid, we're going to start this equation with the HSO3- that was a product of the first equation and dissolve it in water to get SO3 2-.
HSO3- + H2O <----> H3O+ + SO3 2-
Ka2= [H3O] [SO3 2-] / [HSO3-]
6. Make your second ICE table using your initial concentrations from your first ICE table, plug in your equilibrium values into your Ka2 expression, and solve for X.
Note: Ka2 is really small and pretty insignificant when comparing to Ka1. You may want to use the small X approximation here for the -X on the bottom since it is so small that it doesn't make a dent to the number anyway, but either way you get:
X² + .088X -- 5.5x10⁻⁹ for your quadratic. Now solve for X.
X= 6.2x10⁻⁸ (also very small number)
[H3O+] = 0.088 + 6.2x10⁻⁸ = 0.088M
<span>
pH= -log (.088M) = 1.06 or 1.1 for 2 sig figs. </span>
Missing in your question Ka2 =6.3x10^-8 From this reaction: H2SO3 + H2O ↔ H3O+ + HSO3- by using the ICE table : H2SO3 ↔ H3O + HSO3- intial 0.6 0 0 change -X +X +X Equ (0.6-X) X X
when Ka1 = [H3O+][HSO3-]/[H2SO3] So by substitution: 1.5X10^-2 = (X*X) / (0.6-X) by solving this equation for X ∴ X = 0.088 ∴[H2SO3] = 0.6 - 0.088 = 0.512 [HSO3-] = [H3O+] = 0.088
by using the ICE table 2: HSO3- ↔ H3O + SO3- initial 0.088 0.088 0 change -X +X +X Equ (0.088-X) (0.088+X) X
Ka2= [H3O+] [SO3-] / [HSO3-] we can assume [HSO3-] = 0.088 as the value of Ka2 is very small 6.3x10^-8 = (0.088+X)*X / 0.088 X^2 +0.088 X - 5.5x10^-9= 0 by solving this equation for X ∴X= 6.3x10^-8 ∴[H3O+] = 0.088 + 6.3x10^-8 = 0.088 m ( because X is so small) ∴PH= -㏒[H3O+] = -㏒ 0.088 = 1.06
