Calculate the concentrations of all species in a 0.810 M Na2SO3 (sodium sulfite) solution. The ionization constants for sulfurous acid are Ka1 = 1.4 x 10^-2 and Ka2 = 6.3 x 10^-8
I calculated SO32- = 0.8096 M, HSO3- = 3.58E-4, H2SO3 = 7.1E-13, OH- = 3.58E-4, H+ = 2.79E-11
but i keep getting Na+ wrong!!! please help!
I think I answered this late last night that Na^+ = 2 x 0.810M = ?
Is that not right?
Ohh.. I didn't know you were supposed to multiply the SO32- concentration by 2 for Na... That's right thanks DrBob!
Na2SO3 is 100% dissociated; therefore, Na2SO3 ==> 2Na^+ + SO3^2-
So if Na2SO3 is 0.810M than Na^+ is twice that. You are multiplying the Na2SO3 by 2; not multiplying SO3^2- by 2. The SO3^2- is not 0.810 since it hydrolyzes (at least to a slight extent).
What was your second chemical equation that you used?
If you have a question it would be best to post it under "Post a new question."
To calculate the concentration of Na+ in the solution, we need to consider the dissociation of sodium sulfite (Na2SO3) in water.
The balanced equation for the dissociation of Na2SO3 is:
Na2SO3 -> 2Na+ + SO32-
Since Na2SO3 dissociates completely, we know that the concentration of Na+ will be twice that of the initial Na2SO3 concentration.
Given that the initial concentration of Na2SO3 is 0.810 M, we can calculate the concentration of Na+ as follows:
Concentration of Na+ = 2 x Concentration of Na2SO3
= 2 x 0.810 M
= 1.620 M
Therefore, the concentration of Na+ in the solution is 1.620 M.
It's important to note that when calculating concentrations of ionic species, it's crucial to use the stoichiometry of the balanced equation and account for any dissociation or ionization that occurs.