solvolysis of salt of a weak acid and weak base

8/22/2019 Solvolysis of Salt of a Weak Acid and Weak Base

1/11

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:

25ml of 1.75M NaOH with 100ml of 0.5M HCl

Time, t (sec) Temperature, T (C)

60 25.0

120 25.0

180 25.0

240 25.0

309 27.0

313 28.0

320 28.5

348 29.0

408 29.0

468 29.0

528 29.0

588 29.0


2/11

25ml of 1.75M NaOH with 100ml of 0.5M H3BO3


0 25.0

60 25.0

120 25.0

180 25.0

240 25.0

306 26.0

310 26.5

313 27.0

336 27.5

396 27.5

456 27.5

516 27.5

576 27.5


3/11

25ml of 1.75M NH4OH with 100ml of 0.5M HCl


0 25.0

60 25.0

120 25.0

180 25.0

240 25.0

302 25.0

306 26.0

308 26.5

329 27.0

389 27.0

449 27.0

509 27.0

569 27.0


4/11

25ml of 1.75M NH4OH with 100ml of 0.5M H3BO3


0 25.0

60 25.0

120 25.0

180 25.0

240 25.0

307 25.5

313 26.0

330 26.5

390 26.5

450 26.5

510 26.5

570 26.5

830 26.5


5/11

Calculation:

Based on the graphs,

The values of T in mixtures of;

a) 25ml of 1.75M NaOH with 100ml of 0.5M HCl= 4.0C

b) 25ml of 1.75M NaOH with 100ml of 0.5M H3BO3= 2.5C

c) 25ml of 1.75M NH4OH with 100ml of 0.5M HCl= 2.0C

d) 25ml of 1.75M NH4OH with 100ml of 0.5M H3BO3= 1.5 C

The mole of NaOH used is

Mole = MV / 1000

= (1.75M)(25cm3) / 1000

= 0.04375 mol

The mole of HCl used is

Mole = MV / 1000

= (0.50M)(100cm3) / 1000

= 0.05 mol


6/11

Limiting mole is NaOH, hence the enthalpy of neutralisation is as such;

-13.36 k cal mol-1

x 0.04375

= -0.5845 k cal

Heat released = - q

Therefore,

-q1 = msolutionCsolution T + CcalorimeterT

- (-0.5845 x 103

cal) = (125g) (1 cal C-1

g-1

) (4C) + Ccalorimeter(4C)

Ccalorimeter= (584.5cal -500cal) / (4C)

= 21.125 cal C-1

-q4 = (125g) (1 cal C-1

g-1

) (2.5C) + (21.125) (2.5C)

= 365.3 cal

-q5 = (125g) (1 cal C-1

g-1

)(2.0C) + (21.125)(2.0C)

= 292.3 cal

q6 = q4 +q5q1

= (-365.3cal -292.3cal) -(-584.5cal)

= -73.1 cal

Theoretical qis calculated as such;

-q= (125g) (1 cal C-1

g-1

)(1.5C) + (21.125)(1.5C)

= 219.2 cal

Fraction of the salt ammonium borate that reacts with water when dissolved in

water is calculated as such where is the value when the reactants are mixed in

exactly equimolar amount.


7/11

=

= 219.2 cal / 73.1 cal

= 3

The value of is the fraction of the salt ammonium borate that undergoes

solvolysis

= 1 -

= 13

= - 2

Hence the value of equilibrium constant can be identified using the relationship

between Ksolvolysis and , where;

Ksolvolysis =

=

= 4/9

Given that KW is 1 x 10-14

and Kb is 1.75 x 10-5

for ammonia, the value of Ka of

boric acid can be found as shown below;

Ksolvolysis =

Ka = (1 x 10-14

) / ( 4/9 x 1.75 x 10-5

)

= 1.286 x 10-9


8/11

The published value of Ka of boric acid is 7.3 x 10-10

With this, the percentage of error is

x 100%

= 76.2%

Discussion:

Based on the results and calculations, the heat released decreases as the strength

of acidity and basicity reduces. The heat change of each reaction is tabulated in the table

below;

Reactions Heat Released (Exothermic) (cal)

25ml of 1.75M NaOH with 100ml of 0.5M HCl 584.5

25ml of 1.75M NaOH with 100ml of 0.5M H3BO3 365.3

25ml of 1.75M NH4OH with 100ml of 0.5M HCl 292.3

25ml of 1.75M NH4OH with 100ml of 0.5M H3BO3 219.2

Heat evolved during reactions decreases because the reactions are incomplete as the

strength of acid or base decreases. The calculated q6 is higher than the observed value of

q. The reason is that the reaction does not go to completion so the heat that is supposed to

evolve and the observed heat change of reactant is different.

According to the published value of Ka of boric acid, the percentage error is high,

approximately 7.3 x 10-10

. This is owes to external factors such as the type of material

the calorimeter is made up of as well as the efficiency of the calorimeter. Internal factors


9/11

such as the temperature change was not clearly recorded together with the mishandling of

apparatus would also apply to the inaccuracy in the reading.

The results were taken from a group of students in Lab Practical 3 as the results in my

group were inaccurate and there were several discrepancies when the values obtained

were substituted into the calculations above.

Conclusion: 1) The value of calculated Ka of boric acid is 1.286 x 10-9

.

2) The heat capacity of calorimeter is 21.125 cal C-1

.

3) The equilibrium constant for the solvolysis constant is 0.44

Reference: http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch17/ph.php#ka
http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch17/ph.php#kahttp://chemed.chem.purdue.edu/genchem/topicreview/bp/ch17/ph.php#kahttp://chemed.chem.purdue.edu/genchem/topicreview/bp/ch17/ph.php#ka


10/11

Graph:

25ml of 1.75M NaOH with 100ml of 0.5M HCl

25ml of 1.75M NaOH with 100ml of 0.5M H3BO3

24.5

25.0

25.5

26.0

26.5

27.0

27.5

28.0

28.5

29.0

29.5

0 100 200 300 400 500 600 700

temperature(C)

Time, t (seconds)

temperature (C) against time,t

24.5

25.0

25.5

26.0

26.5

27.0

27.5

28.0

0 100 200 300 400 500 600 700

temperature(C)

Time, t (seconds)



11/11

25ml of 1.75M NH4OH with 100ml of 0.5M HCl

25ml of 1.75M NH4OH with 100ml of 0.5M H3BO3

24.5

25.0

25.5

26.0

26.5

27.0

27.5

0 100 200 300 400 500 600

temperature(C)

Time, t (seconds)


24.8

25.0

25.2

25.4

25.6

25.8

26.0

26.2

26.4

26.6

0 100 200 300 400 500 600 700 800 900

temperature(C)

Time, t (seconds)


solvolysis of salt of a weak acid and weak base

Documents