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Acid or Base?

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An assignment for Unit 13, achieved Distinction. All calculations are correct, it has the results for the titrations of all 4 equations, with a graph for each one of them and an explanation of why I chose a certain pH.

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APPLICATIONS OF
INORGANIC CHEMISTRY
ACID OR BASE?
RAFAELA GASPAR ID 192763 BTEC APPLIED SCIENCE

, APPLICATIONS OF INORGANIC CHEMISTRY


TASK 1 & 2
Part A: Calculating pH of solutions of known concentration
Calculate the concentration of [H+] ions and the pH for the following solutions, giving
your answer to 2 decimal places:
1. 0.5M hydrochloric acid (HCl)  [H+] = [HA] = 0.5 mol/dm -3 pH = -log [H+]  pH =
-log (0.5)  pH = 0.30
2. 0.4M nitric acid (HNO3)  [H+] = [HA] = 0.4 mol/dm-3 pH = -log [H+]  pH = -log
(0.4)  pH = 0.40
3. 1.0M sulfuric acid (H2SO4)  [H+] = [HA] x 2 = 2.0 mol/dm -3 pH = -log[H+]  pH =
-log (2.0)  pH = - 0.30
4. 0.2M ethanoic acid  [H+] = Ka x [HA]  [H+] = (1.7 x 10-5) x 0.2  [H+] =
1.84 x 10-3 mol/dm-3 pH = -log [H+]  pH = -log (1.84 x 10-3)  pH = 2.73
5. 0.4M methanoic acid  [H+] = Ka x [HA]  [H+] = (1.8 x 10-4) x 0.4  [H+] =
8.49 x 10-3 mol/dm-3 pH = -log [H+]  pH = -log (8.49 x 10-3)  pH = 2.07
Kw 1 x 10-14
6. 0.1M sodium hydroxide (NaOH)  [H ] =
[ OH- ]  [H ] = 0.1  [H+] = 1 x
+ +


10-13 mol/dm-3 pH = -log [H+]  pH = -log (1 x 10-13)  pH = 13.00
Kw 1 x 10-14
7. 0.2M potassium hydroxide (KOH)  [H+] =  [H+] =  [H+] = 5 x
[ OH- ] 0.2
10-14 mol/dm-3 pH = -log [H+]  pH = -log (5 x 10-14)  pH = 13.30
Kw 1 x 10-14
8. 0.3M calcium hydroxide (Na(OH)2)  [H+] = 2 x M [ OH -]  [H+] = 
2 x 0.3
[H+] = 1.67 x 10-14 mol/dm-3 pH = -log [H+]  pH = -log (1.64 x 10-14)  pH = 13.78
9. A buffer solution of ethanoic acid with sodium ethanoate where the concentration of
the weak acid is 0.200M and the concentration of the conjugate base is 0.122M  -
[A-]
log Ka = 4.77. log = -0.21. pH = 4.77 + (- 0.21)  pH = 4.56
[HA]
10. A buffer solution of phosphoric acid with disodium hydrogen phosphate where the
concentration of weak acid is 0.130M and the concentration of conjugate base is
[A-]
0.026M  -log Ka = 2.15. log = -0.70. pH = 2.15 + (- 0.70)  pH = 1.45
[HA]

Part B: Calculations with a known pH
NB the following calculations will require some rearrangement of the formulae. Give
your answers to 2 decimal places.
1. 0.128M solution of uric acid (HC5H3N4O3) has a pH of 2.39. What is the Ka of this
weak acid? (NB consider uric acid to be a monoprotic acid)  [H+] = 10-pH  [H+] =
[H+]2
10-2.39  [H+] = 4.07 x 10-3 mol/dm-3 [H+] = Ka x [HA]  Ka =  Ka =
[HA]
-3 2
(4.07 x 10 )
 Ka = 1.29 x 10-4 mol/dm-3 at 25C
0.128
2. 0.450M solution of propanoic acid has a pH of 3.68. What is the K a this weak acid?
[H+] = 10-pH  [H+] = 10-3.68  [H+] = 2.09 x 10-4 mol/dm-3 [H+] = Ka x [HA]  Ka
2 -4 2
[H+] (2.09 x 10 )
=  Ka =  Ka = 9.70 x 10-8 mol/dm-3 at 25C
[HA] 0.450
3. A solution of hydrobromic acid has a pH of 1.08 and K a of 1 x 109 mol/dm-3 at 25C.
What is the concentration of acid molecules [HA] in equilibrium in the solution of this


Rafaela Gaspar ID 192763 BTEC Applied Science Level 3

Connected book
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Frances Annets, Shirley Foale, Roy Llewellyn, Ismail Musa, Sue Hocking, BTEC Level 3 National Applied Science, Student Book
Publisher: november 2010 ISBN: 9781846906800 Edition: 3

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