# Do the following titration and find concentration of all species in solution. Diprotic acid H2B is being titrated against 0.16 M NaOH. H2B has the following properties:  40 ml at 0.4 M...

Do the following titration and find concentration of all species in solution. Diprotic acid H2B is being titrated against 0.16 M NaOH. H2B has the following properties:  40 ml at 0.4 M concentration with Ka1 = 2.5 * 10-3 and Ka2 = 3.0 * 10-10.

What  is the concentration of each species in solution at the following points in the titration?

Be sure to clearly list assumptions for each step of this problem.

llltkl | College Teacher | (Level 3) Valedictorian

Posted on

Your question was edited to accord with eNotes policy of not answering multiple questions.

Diprotic acid `H_2B` dissociates in solution in a stepwise manner as:

Step 1. `H_2B stackrel rarr larr H^(+)+HB^-`

Let x be the degree of dissociation for this step. Then concentration of various species in solution are:

` H_2B stackrel rarr larr H^(+)+HB^-`

conc: c(1-x)      cx         cx

`Ka_1 =(cx*cx)/(c(1-x))~~cx^2` (since x is too low compared to 1)

`rArr [H^(+)]=[HB^-]=cx=sqrt(Ka_1*c)`

`=sqrt(2.5*10^(-3)*0.4)=3.16*10^(-2) M`

And `[H_2B]~~c=0.4 M`

Now consider step 2 of the dissociation process.

Step 2. `HB^- stackrel rarr larr H^(+)+B^(2-)`

Let y be the degree of dissociation for this step. Then concentration of various species in solution are:

` H_2B stackrel rarr larr H^(+)+HB^-`

conc:c'(1-y)         c'y     c'y

`Ka_2=(c'y*c'y)/(c'(1-y))~~c'y^2` (since y is too low compared to 1)

`rArr [H^(+)]=[B^(2-)]=c'y=sqrt(Ka_2*c')`

`=sqrt(3.0*10^(-10)*3.16*10^(-2))=3.08*10^(-6) M`

And `[HB^-]~~c'=3.16*10^(-2) M`

Overall, concentration of various species in solution are:

`[H_2B]~=0.4 M`

`[H^(+)]=3.16*10^(-2) M + 3.08*10^(-6) M~~3.16*10^(-2) M`

`[HB^-]=3.16*10^(-2) M`

`[B^(2-)]=3.08*10^(-6)M `

Sources: