How to draw a titration curve for a strong base and strong acid

35.5 mL of a .100M solution of NaOH is added to 35.0 mL of .100M of hydrochloric acid. Draw a titration curve with 1 point in Region 1, 3 points in region 2, 1 point in region 3, and 3 points in region 4. Strong Base and Weak Acid.  

I would guess that you are supposed to calculate the pH of  a weak acid (Region 1), a mixture of the acid and  some additional NaOH (Region 2 or the area I have marked as the buffered area), the equivalence point (where all the acid has just been neutralized - Region 3), and the pH when additional .5 mL of base have been added.


A quick sketch of the curves for a strong acid being titrated by NaOH is shown in black, while the titration curve for a weak acid by NaOH is shown in red.

You might find the tutorial on buffers useful for determining how to calculate the pH in the buffered region.



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Hey thanks. Do you hapen to have the work on how you did it? because I attempted it many times and my Region 3 was at a pH of 3.5 and I do not know what I am doing wrong. Could you also please give me the pH for all the points on the graph so that I may try them and gain my knowledge on how to do it? Thanks again.

Idid not show you a plot of points, but indicated the shape of the titration curves in general.  

Strong acids will maintain a low pH until they are almost at the equivalence point.  They then shoot up almost vertically, and then level off at a high pH.

Weak acids, on the other hand, show a gradual increase in pH as the base is added due to the buffering effect.  Their equivalence points would be above a pH of 7 since they form solutions of basic salts.  

I can't give you exact point for the weak acid vs NaOH without knowing what your weak acid was. Perhaps the work below will help you.

For the strong acid:

Region 1 no base -  .100H HCl - strong acid therefore completely ionized.  [H+] = .1OM, pH =1


Region 2 acid  about half neutralized by adding 17.0 mL of .1M NaOH

                 molarity of acid = (orig. moles of acid - moles of base added) / total volume in liters

                                              = (.0035 moles - .0017mole) / (.035L +.017L) = .035M

              Again, since HCl is a strong acid, [H+] = [HCl] = .035M,    pH= 1.5 ( Note only a slight change in the pH even though the acid is almost half neutralized.  This is typical of titrations of strong acids by strong bases, pH shows very little change until it gets very close to the equivalence/end point)

Region 3 Equivalence Point - equal amounts of base and acid added,  This results in a solution of a neutral salt (NaCl) which has no effect on pH.  The pH of the solution will be the pH of water, 7


Region 4 - Excess base added - .5ml of .1M NaOH has been added (beyond the amount required to neutralize the acid      

                                  molarity of OH- =  moles of excess base/ (volume of acid + base)

                                                            =  (.0005L  x .1M )/ .0705L  = 7.09 x 10-4

                                     pOH =  3.15       

                                      pH = 14 - pOH = 10.9

For a weak acid

Region 1 would require you to know the Ka of the acid and to calculate the concentration of the H+ from the Ka expression.

Region 2 - essentially a buffer (again I would highly recommend you review the tutorial which I posted on this topic.  Sample Problem 2 at the end of the tutorial is almost exactly like this problem.

Region 3 - at the equivalence point you have formed a solution of a basic salt.  The conjugate base of the weak acid will cause the pH of the solution to be somewhat above 7.  You will need to use Kb (which can be determined from the Ka value) and the Kb expression to calculate OH- concentration, pOH, and pH.

Region 4 - essentially a solution of a strong base.  Will be exactly as I showed you for a strong acid.

THANKS SO MUCH. It makes it a lot clearer now.

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