Experiment 8: pH TITRATION OF WEAK The previous experiment demonstrated the appl

Question

Experiment 8: pH TITRATION OF WEAK The previous experiment demonstrated the application of standard titration quantitatively determining the contents of mixtures. An indica titrations to reveal, by means of a color change, when the equivalence point (neutralization point) was reached, that is, when sufficient titrant had been techniques for was used in the with all of the analyte. You will explore in this experiment an alternative a titration, one in which a pH probe and data acquisition system are employed to m the progress of the reaction. approach to Consider the titration of 25 mL of 0.10 MHC,H,O (a weak acid) with 0.10 M strong base): NaOH ( HC110: + NaOH NaCl 1,02 + H2O Because the concentrations of the analyte (the acetic acid) and titrant (the base) are equal, and because the stoichiometry of the analyte and titrant is 1:1, we know that 25 mL of the NaOH solution will be required to reach the equivalence point follow the pH of the acid solution as the NaOH is slowly added to it. The result would be the titration curve shown in Figure 1. The acid solution has a low initial pH (2.87). As . Now suppose that we 14 T 12 10 10 20 mL NaOH 30 40 Figure 1. pH titration curve for titrating 25 mL 0.10 MHC2H,0, with 0.10 M NaOH.

Explanation / Answer

1) Percent error in experimentally determined molar mass:

Molar mass of KHP

HKC8H4O4 = 39.098 + 1.0079 + (12.011*8) + (1.0079*4) + (15.999*4)

= 204.2215

Actual molar mass of KHP = 204.22 g/mol

Experimental molar mass of KHP = 192.31 g/mol (given)

Error = Actual molar mass - Experimental molar mass

204.22 -192.31 = 11.91

Percent error = (Experimental molar mass - Actual molar mass)/ Actual molar mass x 100

Percent error = ((192.31 -204.22)/ 204.22) x 100

Percent error = 5.83%

2a. KHP sample is impure, so less volume of NaOH needed to neutralize it. n = c v

Calculated moles of NaOH are low. 2 moles of NaOH completely neutralizes KHP, so, calculated moles of KHP are also low.

But Ka will remain constant for a particular acid despite a change in concentration

So, there will be no effect on calculated Ka2

PKa2 = - log Ka2

There will be no effect on calculated PKa2

b.

PH = PKa2 + Log [C8H4O42-]/[KHP]

PKa2 = PH - Log [C8H4O42-]/[KHP]

If Log [C8H4O42-]/[KHP] is constant, pKa2 will be more, when pH is more

PKa2 is calculated at pH 4.0 instead of 4.3, so pKa2 will be too low.

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