Titration Curve for a Polyprotic Acid
Titration Curve for a Polyprotic AcidNAME:________________________________________ ¬†PERIOD:___________PrelabShow Calculations.1. For the titration of 20.0 ml of 0.100M¬†maleic acid with 0.100M¬†NaOH, using a Ka1¬†of ¬† ¬† 1.4 x 10-2¬†and Ka2¬†= 8.6×10-7, calculate the pH:a. Initially (0 ml of NaOH added):b. At the first half equivalence point:c. At the first equivalence point:d. At the second half equivalence point:e. At the second equivalence point:f. 10.0 ml beyond the second equivalence point:Titration Curve for a Polyprotic AcidObjectives:¬†¬†¬†¬†¬†¬†¬†¬†In this experiment, a solution of H3PO4¬†will be titrated with a solution of NaOH. The pH of the solution will be monitored as the NaOH is added with a pH probe attached to a CBL. The shape of the pH titration curve will be observed and the Ka¬†values for the acid will be determined.¬†Introduction:¬†¬†¬†¬†¬†¬†¬†¬†Polyprotic acids possess several ionizable hydrogens and undergo stepwise ionization. Each ionization characterized by specific equilibrium constant. Phosphoric acid, H3PO4, is a triprotic acid and shows the following ionization reactions in aqueous solution:H3PO4 (aq¬† ¬†+ ¬† H2O (l)¬† ¬† ‚Üí¬† ¬† H3O+1(aq)¬† ¬† ¬†+ ¬† ¬†H2PO4-1(aq)¬†¬†¬†¬†¬†¬†¬†¬†H2PO4-1¬†(aq¬† ¬†+ ¬† H2O (l)¬† ¬† ‚Üí¬† ¬† H3O+1(aq)¬† ¬† ¬†+ ¬† ¬†HPO4-2(aq)¬†¬†¬†¬†¬†¬†¬†¬†HPO4-2¬†(aq¬† ¬†+ ¬† H2O (l)¬† ¬† ‚Üí¬† ¬† H3O+1(aq)¬† ¬† ¬†+ ¬† ¬†PO4-3(aq)¬†¬†¬†¬†¬†¬†¬†¬†¬†¬†¬†¬†¬†¬†¬†¬†During the titration of H3PO4¬†each hydrogen ion will react will NaOH in a one-to-one ratio with the net ionic equations:H3PO4 (aq)¬† ¬†+ ¬† ¬†OH-1(aq)¬† ¬† ¬†‚Üí¬† ¬† H2O¬†(l)¬† ¬† ¬†+ ¬† ¬†H2PO4-1(aq)H2PO4-1¬†(aq)¬† ¬†+ ¬† ¬†OH-1(aq)¬† ¬† ¬†‚Üí¬† ¬† H2O¬†(l)¬† ¬† ¬†+ ¬† ¬†HPO4-2(aq)HPO4-2¬†(aq)¬† ¬†+ ¬† ¬†OH-1(aq)¬† ¬† ¬†‚Üí¬† ¬† H2O¬†(l)¬† ¬† ¬†+ ¬† ¬†PO4-3(aq)¬†¬†¬†¬†¬†¬†¬†¬†¬†¬†¬†¬†¬†¬†¬†¬†The titration curve should show a jump in pH at each equivalence point as show in the curve for a diprotic acid (Fig.1).[pic 1](Figure 3-Experiment 25-Titration of a Diprotic Acid)- from Chemistry with CBL¬†by Holmquist, Randall, and Volz from Vernier Software 1995)¬†¬†¬†¬†¬†¬†¬†¬†These significant increases in pH at the equivalence points will only be clearly visible if each successive Ka¬†is significantly smaller than the previous one by a factor of 103¬†to 104¬†and the Ka¬†is greater than the ionization constant for water (1×10-14). For instance, if Ka1¬†and Ka2¬†are too close in value, only one jump in pH may be seen at the position of the second equivalence point. If Ka¬†is too small, no jump in pH may be apparent.

If the initial volume of the polyprotic acid and the intial concentrations of the polyprotic acid and NaOH are known, the equivalence points and Ka values can be calculated from pH values on the titration curve.         Ka1 can be calculated using the initial concentration of the acid and the initial pH of the solution. If Ka1 is significantly greater than the other ionization constants by a factor of at least 103, the [H3O+1] can be assumed to come essentially from only the first ionization and can be calculated from the initial pH. The [H2PO4-1] approximately equals [H3O+1] in the initial solution. Since the H3PO4 is only slightly ionized, the [H3PO4] is assumed to be approximately equal to its initial concentration.        Ka1 can be calculated from the pH at the first half-equivalence point. At this point in the titration, half of the moles of H3PO4 have been converted to H2PO4-1. The [H2PO4-1] = [H3PO4], the ratio [H2PO4-1]/[H3PO4] equals one, the [H3O+1] equals Ka1, and the pH of the solution equals pKa1.        Ka2 can be calculated from the pH at the first equivalence point (assuming Ka1 has been calculated). All the moles of H3PO4 have been converted to H2PO4-1. The H2PO4-1 can hydrolyze by the reaction:H2PO4-1 (aq)   +   H2O (l)   →   H3PO4 (aq)    +    OH-1 (aq)It can also ionize further by the reaction:H2PO4-1 (aq   +   H2O (l)    →    H3O+1(aq)     +    HPO4-2(aq)If Ka1 and Ka2 are significantly different, the pH at the first equivalence point will be approximately equal to the average of pKa1 and pKa2.        Ka2 can be calculated from the pH at the second half-equivalence point. At this point in the titration, half of the moles of H2PO4-1 have been converted to HPO4-2. The [HPO4-2] = [H2PO4-1], the ratio [HPO4-2]/[H2PO4-1] equals one, the [H3O+1] equals Ka2, and the pH of the solution equals pKa2.

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Solution Of H3Po4 And H3Po4. (May 31, 2021). Retrieved from https://www.freeessays.education/solution-of-h3po4-and-h3po4-essay/