Calculate the ph of a 0.208 0.208 m solution of ethylenediamine ( h 2 nch 2 ch 2 nh 2 h2nch2ch2nh2). the p k a pka values for the acidic form of ethylenediamine ( h + 3 nch 2 ch 2 nh + 3 h3+nch2ch2nh3+) are 6.848 6.848 ( p k a1 pka1) and 9.928 9.928 ( p k a2 pka2). ph = ph= calculate the concentration of each form of ethylenediamine in this solution at equilibrium.

[ h 2 nch 2 ch 2 nh 2 ] =
[h2nch2ch2nh2]= m m
[ h 2 nch 2 ch 2 nh + 3 ] =
[h2nch2ch2nh3+]= m m
[ h + 3 nch 2 ch 2 nh + 3 ] =
[h3+nch2ch2nh3+]=

pH =11.62:   = 0.204 M;

=

=  

Explanation:

To determine the pH, we have to calculate the final equilibrium concentration for the hydrogen ion by working through both dissociation equilibria of the system. Therefore:

⇔             equation (1)

⇔  equation (2)

We know that:

=

=

Using equation 1 to calculate the value of :

If the equilibrium concentration of = x

Then the equilibrium concentration of = 0.208-x

The equilibrium concentration of = x

Thus:   = x*x/0.208  (if we consider x to be extremely small)

x^2 = *0.208 =

x = =  

The equilibrium concentration of   = 0.208 - = 0.204 M

 = *OH^{-}[/tex]/  

= x* / =x

Thus the equilibrium concentration of   =  

pOH = -log[OH^-] = -log = 2.38

pH = pKw - pOH = 14 -2.38 =11.62

b.final

if you mean hydrogen and oxygen then water is the product