[maccha]

A chemistry problem says that : H2SO3 is dissolved in water and NaOH is then added. Predict the products.

What I'm confused about is H2SO3 being dissolved in water.. would it become HSO3- and then combine to form NaHSO3-? Well I know it wouldn't, in fact, because the answer is NaSO3 but I don't really understand why..

[nj_bartel]

Sulfurous acid has 2 protons that it can lose.

[cliverlong]

A chemistry problem says that : H₂SO₃ is dissolved in water and NaOH is then added. Predict the products.

What I'm confused about is H₂SO₃ being dissolved in water.. would it become HSO₃^- and then combine to form NaHSO₃^-?

I agree with most of what you have written but would make the following slight alteration

H₂SO₃ + H₂O   HSO₃^- + H₃O⁺  (which agrees with what you have written)

HSO₃^- + NaOH + H₃O⁺ NaHSO₃ + 2H₂O (your negative charge on the NaHSO₃ was probably a typo)

Well I know it wouldn't, in fact, because the answer is NaSO₃ but I don't really understand why..

To me that makes no sense since if

HSO₃^- has a charge of -1

Then SO₃^2-, if it exists, must have a charge of 2-, thus yielding the salt, Na₂SO₃. So I agree with you the given answer is wrong (or at the very least I can't understand the formula NaSO₃ for the sulphite salt.)

Now chemguide, http://www.chemguide.co.uk/inorganic/period3/oxidesh2o.html, gives the following:

Sulphur dioxide will also react directly with bases such as sodium hydroxide solution. If sulphur dioxide is bubbled through sodium hydroxide solution, sodium sulphite solution is formed first followed by sodium hydrogensulphite solution when the sulphur dioxide is in excess.

SO₂ + 2NaOH   Na₂SO₃ + H₂O

Na₂SO₃ + H₂O + SO₂   2NaHSO₃

Note:  Sodium sulphite is also called sodium sulphate(IV). Sodium hydrogensulphite is also sodium hydrogensulphate(IV) or sodium bisulphite.


And Online Britannica gives:

When sulfur dioxide is dissolved in water, an acidic solution results. This has long been loosely called a sulfurous acid, H₂SO₃, solution. However, pure anhydrous sulfurous acid has never been isolated or detected, and an aqueous solution of SO₂ contains little, if any, H₂SO₃. Studies of these solutions indicate that the predominant species are hydrated SO₂ molecules, SO₂ · nH₂O. The ions present in these solutions are dependent on concentration, temperature, and pH and include H₃O⁺, HSO³⁻, S₂O₅²⁻, and perhaps SO₃²⁻. However, “sulfurous acid” has two acid dissociation constants. It acts as a moderately strong acid with an apparent ionization of about 25 percent in the first stage and much less in the second stage. These ionizations produce two series of salts—sulfites, containing SO₃²⁻, and hydrogen sulfites, containing HSO³⁻. Only with large cations, such as Rb⁺ (rubidium) or Cs⁺ (cesium), have solid HSO³⁻ salts been isolated. Attempts to isolate these salts with smaller cations tend to yield disulfites as a product of dehydration.
2HSO³⁻   S₂O₅²⁻ + H₂O

[Borek]

However, “sulfurous acid” has two acid dissociation constants. It acts as a moderately strong acid with an apparent ionization of about 25 percent in the first stage and much less in the second stage.

Is this from Britannica? That's stupid, percent ionization is not a reasonable way of giving acid strength. Perhaps they mean 'in saturated solution of SO₂' - but it should be given verbatim.

[cliverlong]

However, “sulfurous acid” has two acid dissociation constants. It acts as a moderately strong acid with an apparent ionization of about 25 percent in the first stage and much less in the second stage.

Is this from Britannica? That's stupid, percent ionization is not a reasonable way of giving acid strength. Perhaps they mean 'in saturated solution of SO₂' - but it should be given verbatim.

Yes, Britannica.

I wasn't so offended by the use of 25% as you. My view was those articles walk the tricky tight rope of being technically accurate but accessible to "the man on the top of the Clapham ominbus" (don't worry, it's a British thing )

So "25% ionization" (to me means)

H₂S0₃  H⁺ + HSO₃^-
(at eqm) 1 - 0.25 = 0.75                     0.25                0.25

gives Ka = (0.25/V)(0.25/V) / (0.75/v) = 0.83V (*) (require V because Ka defined in terms of concentrations)

Rather than the article giving a Ka value and then explaining the meaning of Ka, the author went for the expression "25% ionization". That's my guess, anyway.

What is your view of the rest of the comments in the extract from Britannica?

(*) I can't understand why volume would affect Ka, but that's what the expression tells me - I'm probably making a very basic error here.

Clive

[Borek]

I had to check:

http://en.wikipedia.org/wiki/The_man_on_the_Clapham_omnibus

Percent ionization is concentration dependent - so it is not a reliable measure of acid strength. In theory every infinitely diluted acid is 100% ionized, no matter how weak it is. Give me a Ka and I will tell you at what concentration acid is 25% ionized. While giving Ka value may be an overkill for the Clapham omnibus man, giving him pecent ionization is IMHO worse than not giving any information at all.

(*) I can't understand why volume would affect Ka, but that's what the expression tells me - I'm probably making a very basic error here.

It is not volume. At 25% ionization you have

[H⁺]/C_a = 0.25

so

[H⁺] = 0.25*C_a

and so on. (C_a is analytical concentration). Thus finally Ka will have a concentration units, and that's nothing surprising.

In reality Ka should be unitless, and it is if it is expressed using activities. But don't try to dig here. As we say in Polish - don't touch the sh*.*, it stinks

Otherwise the information seems to be OK.