Question

Using the titration curve collected in lab for maleic acid: a. (4 pts) Clearly draw vertical and horizontal lines at the appropriate location on your printed graph where pkai and paz can be estimated. Take a picture of the modified graph with the added pk, lines and insert the image into your assignment submission. b. (4 pts) Using your graph, determine the estimated pkai and pKaz values and then calculate Kal and Kaz.

Please help with parts a and b and show work if possible. Thanks!

Answer 1

(a) At any equivalence point sharp change in pH occurs. That is magnitude of d(pH)/dV is highest at the equivalence point.

For the given titration curve this occurs first at V=10.5 ml and then at 21.05 ml, thus first equivalence occurs at V = 10.5 ml of base added and second equivalence point at V= 21.05 ml of base added.

Lets write maleic as H2A for convenience.

Now when we are at halfway to the first equivalence point then [H2A]= [HA^-], so Ka1= [HA^-]x[H⁺]/[H2A] gives Ka1 = [H⁺] and so -logKa1= -log[H+] and hence pKa₁ = pH at midway to first equivalence point that is when volume of base added is V=10.5/2 ml= 5.25 ml. Similarly at midway point between first and second equivalence point (when V of base added is 10.5+(21.05-10.5)/2 =15.75 ml), we have pKa₂ = pH.

Now we find these two points on the plot as follows:

Point where pka2=pH 6.19 Point where pkai Ерн 2.00 .00000 0 Volume tml)

(b) As found above in the plot we get pKa1 =2.00 and pKa2= 6.19

Now since pKa = -logKa it gives Ka=10^-pKa , thus Ka₁ = 10^2.00 = 1.00x10² and Ka₂ = 10^6.19 = 1.55x10⁶.

Comment in case of any doubt.