We want to calculate the concentrations of all species in a 0.58 M Na 2 SO 3 (sodium sulfite) solution. The ionization constants for sulfurous acid are K a1 = 1.4 × 10 − 2 and K a2 = 6.3 × 10 − 8 . Let's start by first calculating the concentration of the spectator ion.

Explanation:

Reaction equation is as follows.

Here, 1 mole of produces 2 moles of cations.

                                  = 1.16 M

= 0.58 M

The sulphite anion will act as a base and react with to form and .

As,    

                       =

                       =

According to the ICE table for the given reaction,

Initial:        0.58             0              0

Change:     -x               +x             +x

Equilibrium: 0.58 - x     x               x

So,

                x = 0.0003 M

So,   x = = 0.0003 M

= 0.58 - 0.0003

                     = 0.579 M

Now, we will use = 0.0003 M

The reaction will be as follows.

Initial:   0.0003

Equilibrium:  0.0003 - x        x             x

                      =

                      =

Therefore,  

As,  x <<<< 0.0003. So, we can neglect x.

Therefore,  

                              =

                     x =

x = =

                =

                = M

Thus, we can conclude that the concentration of spectator ion is M.

gas

explanation:

hdahhd

              nitroswawdasdwasdawsdwasdwasdwasdwaswasdwasdawdswasdwadwadswadadadasd gay