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

Initial temperature of Ammonia solution = 25.0

Initial temperature of HCl = 24.0

Table 3:

From the graph of Temperature of mixture against Time,

ΔT3 = 4.5°C

Part IV: 25ml of 1.75M Ammonia solution + 100ml of 0.5M H3BO3

Initial temperature of Ammonia solution = 25.0

Initial temperature of H3BO3 = 25.0

Table 4:

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

• Number of mole of NaOH is smaller than concentration of HCl, thus, NaOH is the limiting agent.
• 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.
• Part I: Limiting reagent is sodium hydroxide, NaOH solution.
• Part II: Limiting reagent is sodium hydroxide, NaOH solution
• Part III: Limiting reagent is ammonia, NH3 solution
• Part IV: Limiting reagent is ammonia, NH3 solution

(ii) Determination of heat capacity of calorimeter, Ccalorimeter

Given that for a strong acid-strong base neutralization,

        

                Eq(1)

Assume that:

Specific heat capacity of calorimeter = 1.0 cal °C-1 g-1

Density of solution = 1g cm-3 (125ml solution = 125g solution)

 

Part II: 25ml of 1.75M NaOH + 100ml of 0.5M H3BO3 _        

The equation for the weak acid-strong base neutralization is:

H3BO3 (aq) + OH- (aq)  H2BO3- (aq) + H2O (l)    

        Eq(2)

The limiting reagent is NaOH, n= 0.04375mol

Based on part I, Ccalorimeter = 42.0cal °C-1, ΔT2 = 3.5°C

         

Part III: 25ml of 1.75M Ammonia solution + 100ml of 0.5M HCl                

The equation for the strong acid-weak base neutralization is:

NH3 (aq) + H+ (aq)  NH4+ (aq)                

                Eq(3)

 

The limiting reagent is Ammonia solution, n = 0.04375mol

Ccalorimeter = 42.0 cal °C-1, ΔT3 = 4.5°C

         

Part IV: 25ml of 1.75M Ammonia solution + 100ml of 0.5M H3BO3

The equation for the weak acid-weak base neutralization is:

H3BO3 (aq) + NH3 (aq)  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)  H2BO3- (aq) + H2O (l)    

        Eq(2)

NH3 (aq) + H+ (aq)  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)  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, 2.282

does not deviate much from the theoretical value.