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Question:Titrating diprotic acid with strong base.

Prepare a family of graphs for the titration of 50.0 mL of 0.020 0 M H2A with 0.100 M NaOH. Consider the following cases: (a) pK1 = 4.00, pK2 =8.00; (b) pK1 = 4.00, pK2 = 6.00; (c) pK1 = 4.00, pK2 = 5.00

Short Answer

Expert verified

(b) The plot of the for pK1 = 4.00, pK2 =6.00 is shown below

Step by step solution

01

Information given

50.0 mL of 0.020 0 M H2A needs to be titrated with 0.100 M NaOH

. (pK1=4.00, pK2=6.00)

02

Equation need to be used to develop spreadsheet

Fromtable 11.5 the equation for the fraction of titration of diprotic acid (H2A) with strong base (B) is as follows

ϕ=CbVbCaVa=αHA-+2αA2--H+-OH-Ca1+H+-OH-Cb

Other formulas need to be used

H+=10-pHOH-=KwH+αHA-=H+K1H+2+H+K1+K1K2αA2-=K1K2H+2+H+K1+K1K2

03

Spreadsheet to plot curve

Spreadsheet forpK1=4.00, pK2=6.00

04

Final plot

From the above spreadsheet the following curve was plotted

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Most popular questions from this chapter

Effect of pKb in the titration of weak base with strong acid.Using the appropriate equation in Table 11-5, compute and plot a family of curves analogous to the left part of Figure 11-3 for the titration of 50.0 mL of 0.020 0 M B (pKb = -2.00, 2.00, 4.00, 6.00, 8.00, and 10.00) with 0.100 M HCl. (The value pKb = -2.00 represents a strong base.) In the expression forαBH+,KBH+=KwKb

11-20. The graph shows the titration curve for a protein containingamino acids with 16 basic and acidic substituents. The curve is smooth without clear breaks because 29 groups are titrated in thepH interval shown. The 29 endpoints are so close together that a nearly uniform rise results. The isoionic point is thepHof the pure protein with no ions present exceptH+ and OH-. The isoelectric point is the pHat which the average charge on the protein is zero. Is the average charge of the protein-positive, negative, or neutral at its isoionic point? How do you know?

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Titration on Diprotic Systems

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(a) Calculate the pH at the second equivalence point.

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Consider the titration of 50.0mL of 0.0500M malonic acid with 0.100MNaOH. Calculate the pH at each point listed and sketch the titration curve:Vb=0.0,8.0,12.5,19.3,25.0,37.5,50.0and 56.3 mL.

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