Objective:

1. To determine the extent of solvolysis of ammonium borate in water by calorimeter

Results and calculation:

Part I: Mixture of sodium hydroxide, NaOH and hydrochloric acid, HCl

**Table 1.1 The temperature of mixture and the time taken from beginning**

Time taken from beginning, t (s) | Temperature, T (°C) |

0.0 | 26.0 |

0.5 | 27.0 |

3.5 | 28.0 |

5.5 | 28.5 |

7.5 | 29.0 |

10.5 | 29.0 |

12.0 | 29.5 |

14.5 | 29.5 |

16.5 | 29.5 |

19.5 | 30.0 |

22.0 | 30.0 |

25.0 | 30.0 |

29.5 | 30.0 |

32.0 | 30.0 |

34.0 | 30.0 |

94.0 | 30.0 |

154.0 | 30.0 |

214.0 | 30.0 |

274.0 | 30.0 |

324.0 | 30.0 |

**Graph 1 Temperature of mixture against time taken from beginning**

**Table 1.2 Temperature change of mixture caused by neutralization reaction**

Initial temperature, T | 26.0 |

Final temperature, T | 30.0 |

Change of temperature, ΔT (°C) | 4.0 |

Part II: Mixture of sodium hydroxide, NaOH and boric acid, H_{3}BO_{3}

**Table 2.1 The temperature of mixture and the time taken from beginning**

Time taken from beginning, t (s) | Temperature, T (°C) |

0.0 | 26.0 |

5.5 | 27.0 |

8.5 | 28.0 |

13.0 | 28.0 |

14.0 | 28.0 |

15.0 | 28.0 |

24.0 | 28.5 |

32.0 | 29.0 |

44.5 | 29.0 |

58.0 | 29.0 |

62.0 | 29.0 |

69.0 | 29.0 |

129.0 | 29.0 |

189.0 | 29.0 |

249.0 | 29.0 |

309.0 | 29.0 |

369.0 | 29.0 |

**Graph 2 Temperature of mixture against time taken from beginning**

**Table 2.2 Temperature change of mixture caused by neutralization reaction**

Initial temperature, T | 26.0 |

Final temperature, T | 29.0 |

Change of temperature, ΔT (°C) | 3.0 |

Part III: Mixture of ammonia, NH_{4}OH and hydrochloric acid, HCl

**Table 3.1 The temperature of mixture and the time taken from beginning**

Time taken from beginning, t (s) | Temperature, T (°C) |

0.0 | 25.0 |

2.0 | 26.5 |

5.0 | 27.0 |

6.0 | 28.5 |

8.0 | 29.0 |

10.5 | 29.0 |

12.5 | 29.0 |

15.0 | 29.0 |

17.5 | 29.5 |

20.0 | 29.5 |

23.0 | 29.5 |

26.0 | 30.0 |

29.5 | 30.0 |

33.0 | 30.0 |

39.0 | 30.0 |

99.0 | 30.0 |

159.0 | 30.0 |

219.0 | 30.0 |

279.0 | 30.0 |

339.0 | 30.0 |

**Graph 3 Temperature of mixture against time taken from beginning **

**Table 3.2 Temperature change of mixture caused by neutralization reaction**

Initial temperature, T | 25.0 |

Final temperature, T | 30.0 |

Change of temperature, ΔT (°C) | 5.0 |

Part IV: Mixture of ammonia, NH_{4}OH and boric acid, H_{3}BO_{3}

**Table 4.1 The temperature of mixture and the time taken from beginning**

Time taken from beginning, t (s) | Temperature, T (°C) |

0.0 | 25.0 |

3.5 | 26.0 |

7.0 | 27.0 |

11.5 | 27.0 |

15.5 | 27.0 |

18.0 | 27.0 |

22.5 | 27.0 |

27.5 | 27.0 |

33.0 | 27.0 |

37.5 | 27.0 |

42.0 | 27.0 |

47.5 | 27.0 |

52.5 | 27.0 |

88.0 | 27.0 |

148.0 | 27.0 |

208.0 | 27.0 |

268.0 | 27.0 |

328.0 | 27.0 |

**Graph 4 Temperature of mixture against time taken from beginning **

Initial temperature, T | 25.0 |

Final temperature, T | 27.0 |

Change of temperature, ΔT (°C) | 2.0 |

__Determination of limiting agent in each part__

In the reaction for part I, the limiting agent is calculated as shown below:

Therefore, limiting agent in the reaction is NaOH solution.

The limiting agent in each part is determined by using the method as shown in Part I.

__Determination of heat capacity of calorimeter__

From the enthalpy change, it can be expressed in the term of

-q = (m_{solution})(c_{solution})(ΔT) + (c_{calorimeter} )( ΔT)

-(-ΔH )(n) = (m_{solution})(c_{solution})(ΔT) + (c_{calorimeter} )( ΔT)

Since -ΔH = -q / n, and ΔH is negative because neutralization is an exothermic reaction.

**Assumptions:**

**Specific heat capacity of calorimeter = 1.0 cal °C ^{-1} g^{-1}**

**Density of solution = 1g cm ^{-3} (125ml solution = 125g solution)**

__Part I__

In part I, the enthalpy change of a strong base and a strong acid is -13.36 kcal mol^{-1}.

Number of moles of reaction = 0.04375 mol since limiting agent has 0.04375 mol.

Let ΔH_{1 }be the enthalpy change of neutralization reaction of part I.

-(-ΔH_{1} )(n) = (m_{solution})(c_{solution})(ΔT) + (c_{calorimeter} )( ΔT)

-(-13.36x10^{3} cal mol^{-1})(0.04375mol) = (125g)( 1.0 cal °C^{-1} g^{-1})(4°C) + (c_{calorimeter} )(4°C)

c_{calorimeter }= 21.125g °C^{-1}

_{}__Determination of enthalpy change of each reaction__

__Part II__

Based on part I, c_{calorimeter }= 21.125 cal °C^{-1}, ΔT = 3°C

Number of moles of reaction = 0.04375 mol since limiting agent has 0.04375 mol.

Let ΔH_{2 }be the enthalpy change of neutralization reaction of part II.

-(-ΔH_{2} )(n) = (m_{solution})(c_{solution})(ΔT) + (c_{calorimeter} )( ΔT)

-(-ΔH_{2})(0.04375mol) = (125g)( 1.0 cal °C^{-1} g^{-1})(3°C) + (21.125 cal °C^{-1})(3°C)

ΔH_{2 }= 10.02 kcal mol^{-1}^{}

Since neutralization is an exothermic process, so ΔH_{2} = -10.02 kcal mol^{-1}.

__ __

__ __

__Part III__

Based on part I, c_{calorimeter }= 21.125 cal °C^{-1}, ΔT = 5°C

Number of moles of reaction = 0.04375 mol since limiting agent has 0.04375 mol.

Let ΔH_{3 }be the enthalpy change of neutralization reaction of part III.

-(-ΔH_{3} )(n) = (m_{solution})(c_{solution})(ΔT) + (c_{calorimeter} )( ΔT)

-(- ΔH_{3})(0.04375mol) = (125g)( 1.0 cal °C^{-1} g^{-1})(5°C) + (21.125 cal °C^{-1})(5°C)

ΔH_{3 }= 16.7 kcal mol^{-1}^{}

Since neutralization is an exothermic process, so ΔH_{3} = -16.7 kcal mol^{-1}. However, this value is not valid due to certain error. The enthalpy change of the neutralization reaction between a weak base and a strong acid should be less than the reaction between strong acid and strong base.

__Part IV__

Based on part I, c_{calorimeter }= 21.125 cal °C^{-1}, ΔT = 2°C

Number of moles of reaction = 0.04375 mol since limiting agent has 0.04375 mol.

Let ΔH_{4 }be the enthalpy change of neutralization reaction of part IV.

-(-ΔH_{4} )(n) = (m_{solution})(c_{solution})(ΔT) + (c_{calorimeter} )( ΔT)

-(- ΔH_{4})(0.04375mol) = (125g)( 1.0 cal °C^{-1} g^{-1})(2°C) + (21.125 cal °C^{-1})(2°C)

ΔH_{4 }= 6.68 kcal mol^{-1}^{}

Since neutralization is an exothermic process, so ΔH_{4} = -6.68 kcal mol^{-1}.

__Determination of enthalpy change of dissociation of H _{3}BO_{3} and NH_{3}__

**Assumption: The reaction between weak base and weak acid goes to completion. **

By using the ΔH’s value of obtained,

ΔH_{4} = ΔH_{2} + ΔH_{3} – ΔH_{1}

ΔH_{4} = -10.02 kcal mol^{-1 }+ (-16.7 kcal mol^{-1}) – (-13.36kcal mol^{-1})

ΔH_{4} = -13.36 kcal mol^{-1}

The enthalpy change of neutralization between a weak base and a weak acid is -13.36 kcal mol^{-1}. By comparing to the value calculated, ΔH_{4} = -6.68 kcal mol^{-1} (from part IV), this value is bigger. According to the reaction between weak base and weak acid, this value is not valid because the expected value should be less than the value obtained in part I, part II and part III. Since the value of ΔH_{3} is not valid, therefore the value obtained by using ΔH_{4} = ΔH_{2} + ΔH_{3} – ΔH_{1} is no longer valid.

__Determination of fraction of the salt ammonium borate reacts with water__

Let a be the fraction of ammonium borate salt reacts with water

K_{solv} = equilibrium constant of solvolysis

a = 1 – a’ = 1 – (ΔH / ΔH_{4})

K = a^{2} / (1-a)^{2}

Based on the formula given above, the fraction of ammonium borate salt reacts with water and equilibrium constant of solvolysis were calculated in each part.

For ΔH = -16.7 kcal mol^{-1}, the value of fraction of the salt ammonium borate reacts with water is negative value which is -0.25. This is an invalid value. Hence, the value of equilibrium constant of solvolysis is not valid too.

__Determination of the dissociation constant, K _{a} for boric acid__

From the previous part, the equilibrium constant of solvolysis for part IV is 1.00, and the value of K_{w} = 1.0 x 10^{-14} and K_{b} = 1.75 x 10^{-5}. Therefore, the value of K_{a} of boric acid is calculated as shown below:

1.00 = 1.0 x 10^{-14} / (K_{a})(1.75 x 10^{-5})

K_{a} = 5.71 x 10^{-10}

According to the journal article, Perelygin, Yu. P. & D. Yu. Chistyakov (2006). Boric acid. *Russian Journal of Applied Chemistry,* *79(12),* 2041-2042. doi: 10.1134/S1070427206120305. (Dissociation constant of K_{a} is 7.3 x 10^{-10})

The value obtained experimentally is 5.71 x 10^{-10} which shows a deviation with the value of K_{a} obtained via journal article.

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