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.