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?

  1. 0.00 ml base added

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

Asked on by sethpolk92

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llltkl | College Teacher | (Level 3) Valedictorian

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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:

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