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# Objective: To determine the extent of solvolysis of ammonium borate in water by calorimetry.

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Introduction

ï»¿Title : Solvolysis of the Salt of A Weak Acid and A Weak Base Objective: To determine the extent of solvolysis of ammonium borate in water by calorimetry. Results: Part I: 25ml of 1.75M NaOH + 100ml of 0.5M HCl Initial temperature of NaOH = 27.0 Initial temperature of HCl = 26.0 Table 1: Time(seconds) Temperature of mixture() Time(seconds) Temperature of mixture() 0.0 26.5 360.0 29.5 30.0 29.0 420.0 29.5 60.0 30.0 480.0 29.5 90.0 30.0 540.0 29.5 120. 30.0 600.0 29.5 150.0 29.5 660.0 29.5 180.0 29.5 720.0 29.5 210.0 29.5 780.0 29.5 240.0 29.5 840.0 29.5 270.0 29.5 900.0 29.5 300.0 29.5 From the graph of Temperature of mixture against Time, ΔT1 = 3.5°C Part II: 25ml of 1.75M NaOH + 100ml of 0.5M H3BO3 Initial temperature of NaOH = 25 Initial temperature of H3BO3 = 25.0 Table 2: Time(seconds) Temperature of mixture() Time(seconds) Temperature of mixture() 0.0 25.0 360.0 28.0 30.0 28.5 420.0 28.0 60.0 28.0 480.0 28.0 90.0 28.0 540.0 28.0 120. ...read more.

Middle

28.0 600.0 28.0 150.0 28.0 660.0 28.0 180.0 28.0 720.0 28.0 210.0 28.0 780.0 28.0 240.0 28.0 840.0 28.0 270.0 28.0 900.0 28.0 300.0 28.0 From the graph of Temperature of mixture against Time, ΔT4 = 3.0°C Calculation: Part I: 25ml of 1.75M NaOH + 100ml of 0.5M HCl (i)Determination the limiting reagent To find concentration of NaOH, M1V1 = M2V2 (1.75 M)(25ml) = (M2) (125ml) M NaOH = 0.35 M Moles of NaOH = MV /1000 = (0.35 M)(125 ml) / 1000 = 0.04375 mol To find concentration of HCl, M1V1 = M2V2 (0.5 M)(100ml) = (M2)(125ml) MHCl = 0.40 M Moles of HCl = MV /1000 = (0.40 M)(125 ml) / 1000 = 0.05 mol 1. Number of mole of NaOH is smaller than concentration of HCl, thus, NaOH is the limiting agent. 2. Since volume and concentration for the base and the acid is the same as part I, the limiting reagent for each subpart of the experiment is the base. 3. Part I: Limiting reagent is sodium hydroxide, NaOH solution. 4. ...read more.

Conclusion

+ NH3 (aq) ï H2BO3- (aq) + NH4+ (aq) The limiting reagent is Ammonia solution, n = 0.04375mol Ccalorimeter = 42.0 cal °C-1, ΔT4 = 3.0°C Determination of equilibrium constant of boric acid H3BO3 (aq) + OH- (aq) ï H2BO3- (aq) + H2O (l) Eq(2) NH3 (aq) + H+ (aq) ï NH4+ (aq) Eq(3) Eq(1) H3BO3 (aq) + NH3 (aq) H2BO3- (aq) + NH4+ (aq) = [13.36 + (-13.36) + (-17.18)] kcalmol-1 = -17.18 kcalmol-1 Solvolysis occurs when the neutralization process between boric acid and ammonia solution proceed in the reverse direction: H2BO3- (aq) + NH4+ (aq) ï H3BO3 (aq) + NH3 (aq) The fraction of Given that: , (for ammonia) = 2.282 From Bettelheim, F. A., Brown, W. H., Campbell, M. K., & Farrell, S. O. (2009). Chapter 8: Acids and Bases. In Introduction to General, Organic, and Biochemistry (9th ed., p. 249). Brooks/Cole, Cengage Learning, the Ka for boric acid is 7.3. The Ka value obtained, 2.282does not deviate much from the theoretical value. ...read more.

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