To create a theoretical titration curve for the neutralization of a weak acid by

Question

To create a theoretical titration curve for the neutralization of a weak acid by a strong base. Complete the Table shown below by calculating the missing information. (Be sure to provide example calculations that illustrate clear understanding of the mathematical operations applied to the Table.) Create a plot by graphing pH (y-axis) versus Volume (mL)_HaOH (x-axis). Use graph paper to hand draw the graph or use Microsoft Excel to create plot and attach printout of plot. Then, create a second plot by graphing Delta pH/Delta Volume_NAOH (y-axis) versus mL NaOH_added (x-axis) This rate change plot is an alternative means to identify the Equivalent Point for the titration. Titration of 40.0 mL of 0.100M HCI (strong acid) with 0.100M NaOH (strong base). Example calculations (use additional pages to show calculations of key points of graph):

Explanation / Answer

Titration of HCl with NaOH

NaOH = 0 ml

pH = -log[H+] = -log(0.1) = 1

....

NaOH = 1.25 ml

[H3O+] remained = (0.1 x 40 - 0.1 x 1.25)/41.25 = 0.094 M

pH = -log[H+] = -log(0.094) = 1.03

.....

NaOH = 2.50 ml

[H3O+] remained = (0.1 x 40 - 0.1 x 2.50)/42.5 = 0.088 M

pH = -log[H+] = -log(0.088) = 1.05

.....

NaOH = 5.00 ml

[H3O+] remained = (0.1 x 40 - 0.1 x 5)/45 = 0.078 M

pH = -log[H+] = -log(0.078) = 1.109

.....

NaOH = 7.5 ml

[H3O+] remained = (0.1 x 40 - 0.1 x 7.5)/47.5 = 0.068 M

pH = -log[H+] = -log(0.068) = 1.165

.....

NaOH = 10 ml

[H3O+] remained = (0.1 x 40 - 0.1 x 10)/50 = 0.06 M

pH = -log[H+] = -log(0.06) = 1.222

.....

NaOH = 15 ml

[H3O+] remained = (0.1 x 40 - 0.1 x 15)/55 = 0.045 M

pH = -log[H+] = -log(0.045) = 1.34

.....

NaOH = 20 ml

[H3O+] remained = (0.1 x 40 - 0.1 x 20)/60 = 0.033 M

pH = -log[H+] = -log(0.033) = 1.48

.....

NaOH = 21 ml

[H3O+] remained = (0.1 x 40 - 0.1 x 21)/61 = 0.031 M

pH = -log[H+] = -log(0.031) = 1.506

.....

NaOH = 22.5 ml

[H3O+] remained = (0.1 x 40 - 0.1 x 22.5)/62.5 = 0.028 M

pH = -log[H+] = -log(0.028) = 1.55

.....

NaOH = 25 ml

[H3O+] remained = (0.1 x 40 - 0.1 x 25)/65 = 0.0231 M

pH = -log[H+] = -log(0.0231) = 1.637

.....

NaOH = 27.5 ml

[H3O+] remained = (0.1 x 40 - 0.1 x 27.5)/67.5 = 0.0185 M

pH = -log[H+] = -log(0.0185) = 1.73

.....

NaOH = 30 ml

[H3O+] remained = (0.1 x 40 - 0.1 x 30)/70 = 0.0143 M

pH = -log[H+] = -log(0.0143) = 1.84

.....

NaOH = 32.5 ml

[H3O+] remained = (0.1 x 40 - 0.1 x 32.5)/72.5 = 0.0103 M

pH = -log[H+] = -log(0.0103) = 1.985

.....

NaOH = 35 ml

[H3O+] remained = (0.1 x 40 - 0.1 x 35)/75 = 0.0067 M

pH = -log[H+] = -log(0.0067) = 2.176

.....

NaOH = 37.5 ml

[H3O+] remained = (0.1 x 40 - 0.1 x 37.5)/77.5 = 0.0032 M

pH = -log[H+] = -log(0.0032) = 2.49

.....

NaOH = 40 ml

Neutralizarion point

pH = 7

.....

NaOH = 45 ml

[OH-] remained = (0.1 x 45 - 0.1 x 40)/85 = 0.00588 M

pH = 14 - log[OH-] = 14-log(0.00588) = 11.77

.....

NaOH = 50 ml

[OH-] remained = (0.1 x 50 - 0.1 x 40)/90 = 0.011 M

pH = 14 - log[OH-] = -log(0.011) = 12.04

.....

NaOH = 70 ml

[OH-] remained = (0.1 x 70 - 0.1 x 40)/110 = 0.0273 M

pH = 14 - log[OH-] = -log(0.0273) = 12.436

.....

NaOH = 80 ml

[OH-] remained = (0.1 x 80 - 0.1 x 40)/120 = 0.033 M

pH = 14 - log[OH-] = -log(0.033) = 12.523

.....

Plot the data

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