Posted by sk1227 on Thursday, April 14, 2011 at 8:17am.
I have no idea what concn is being titrated in Fig 17-9; however, if we consider it 100 mL of 0.1M HCl being titrated with 0.1M NaOH, we can ESTIMATE it this way.
You start with 0.1M and you want the concn of H^+ to be 0.01 (pH of 2) at the end; therefore, the concn must be reduced by about 0.09. We can make an approximate correction for the dilution by the titrate then as 0.09 M x (100/109 = about 0.0826M.
If we start with 0.1M we will still have 0.1M-0.0826M = 0.0174 M at pH = 2 and the percent is (0.0174/0.1)*100 = about 17%. That is an estimate only. The actual percent, for 0.1M, can be calculated as follows:
Assume we take 100 mL of 0.1M HCl (10 mmoles) and titrate with 0.1M NaOH. Let x = mL NaOH added.
([mmoles HCl - 0.1x)/(mL HCl + x)] = 0.01M
(10-0.1x) = (0.01)(100+x) and solve for x. If I didn't make a mistake x = 81.8 and we check to see if that is a good value.
10 mmoles HCl to start
-8.18 mmoles NaOH added
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1.82 mmoles HCl left untitrated.
(1.82 mmoles/10 initially)*100 = 18.2% not yet titrated so a significant error is made by using thymol blue. This percent will change as the concn changes; therefore, it your Figure 17-9 is not 0.1M, a different percent of HCl will remain untitrated at pH = 2.