# Concentrations of acids and bases?

## Calculate the concentrations of all species in a 1.70 M Na2SO3 (sodium sulfite) solution. The ionization constants for sulfurous acid are Ka1 = 1.4× 10–2 and Ka2 = 6.3× 10–8. Na+ SO3^2- HSO3^- H2SO3 OH- H+ Any help would be appreciated

Mar 9, 2018

See Below

#### Explanation:

Na_2SO_3 ↔ 2Na^+ + SO_3^"-2"
[Na+] = 2x1.7M = 3.4M

Now you have $S {O}_{3}^{\text{-2}}$ in Water, or a weak base in Water.
$S {O}_{3}^{\text{-2" + H_2O ↔ HSO_3^"-1}} + O {H}^{-}$
You are given the Ka values, but for this instance, you need the Kb (using Ka2)
${K}_{A} \times {K}_{B} = {K}_{W}$
${K}_{B} = 1.587 \times {10}^{\text{-7}}$
ICE table
$S {O}_{3}^{\text{-2" + H_2O ↔ HSO_3^"-1}} + O {H}^{-}$
I 1.7 0 0
C -x +x +x
E 1.7(assumption) +x +x
${K}_{B} = {x}^{2} / \text{1.7M}$
x = 0.000519
pOH=3.28; [OH^-] = 5.25xx10^-4M
pH=10.71; [H^+] = 1.95xx10^-11M

The next step is sort of strange. I don't really know any better way to determine this other than using the Henderson Hasselbalch equation (don't want to break out my Quantitative Analysis book)

I'm going to pretend that ${H}_{2} S {O}_{3} \mathmr{and} H S {O}_{3}^{-}$ are going to make a buffer. the pKa of this buffer system is 1.85, but the pH = 10.71!! This is the pH from above, and I'm using it to figure out how much ${H}_{2} S {O}_{3}$ might be in solution (theoretically there will be some, since it asymptotically approaches zero as the pH goes up...but it will be extremely small.

$p H = p K a + \log \left(\frac{{H}_{S} O {3}^{-}}{{H}_{2} S {O}_{3}}\right)$
$10.71 = 1.85 + \log \left(\frac{0.000519}{{H}_{2} S {O}_{3}}\right)$
${10}^{8.86} = \left(\frac{0.000519}{{H}_{2} S {O}_{3}}\right)$
$\left[{H}_{2} S {O}_{3}\right] = 7.16 \times {10}^{-} 13 M$
This is obviously a very small amount of ${H}_{2} S {O}_{3}$, and you can't even substract it from the concentrations above, since you don't have 13 significant figures for these calculations. So you can call it zero, but your teacher/professor might want to see you try to make the attempt at calculating it, so there it is.

All the Parts
$\left[N {a}^{+}\right] = 2 \times 1.7 M = 3.4 M$
$\left[S {O}_{3}^{-} 2\right] = 1.7 - 0.000519 = 1.699481 M$ (too many sig figs)
$\left[H S {O}_{3}^{-}\right] = 0.000519 M$
$\left[{H}_{2} S {O}_{3}\right] = 7.16 \times {10}^{-} 13 M$ (or zero)
$\left[O {H}^{-}\right] = 5.25 \times {10}^{-} 4 M$
$\left[{H}^{+}\right] = 1.95 \times {10}^{-} 11 M$