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

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

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Dave Share
Mar 9, 2018

Answer:

See Below

Explanation:

Ok, I'll start with the easy one.
#Na_2SO_3 ↔ 2Na^+ + SO_3^"-2"#
[Na+] = 2x1.7M = 3.4M

Now you have #SO_3^"-2"# in Water, or a weak base in Water.
#SO_3^"-2" + H_2O ↔ HSO_3^"-1" + OH^-#
You are given the Ka values, but for this instance, you need the Kb (using Ka2)
#K_AxxK_B = K_W#
#K_B = 1.587xx10^"-7"#
ICE table
#SO_3^"-2" + H_2O ↔ HSO_3^"-1" + OH^-#
I 1.7 0 0
C -x +x +x
E 1.7(assumption) +x +x
#K_B = x^2/"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_2SO_3 and HSO_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_2SO_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.

#pH = pKa + log((H_SO3^-)/(H_2SO_3))#
#10.71 = 1.85 + log((0.000519)/(H_2SO_3))#
#10^8.86 = ((0.000519)/(H_2SO_3))#
#[H_2SO_3] = 7.16xx10^-13 M#
This is obviously a very small amount of #H_2SO_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
#[Na^+] = 2xx1.7M = 3.4M#
#[SO_3^-2] = 1.7-0.000519 = 1.699481 M# (too many sig figs)
#[HSO_3^-] = 0.000519 M#
#[H_2SO_3] = 7.16xx10^-13 M# (or zero)
#[OH^-] = 5.25xx10^-4M#
#[H^+] = 1.95xx10^-11M#

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