BTEC Level 3 National Applied Science, Student Book
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.
BTEC Applied Science Unit 2, Assignment D (FULL ASSIGNMENT)
BTEC Applied Science Unit 2, Assignment C (FULL ASSIGNMENT)
BTEC Applied Science Unit 2, Assignment B (FULL ASSIGNMENT)
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PEARSON (PEARSON)
Applied Science 2016 NQF
Unit 13 - Applications of Inorganic Chemistry
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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 25C
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 25C
[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 25C.
What is the concentration of acid molecules [HA] in equilibrium in the solution of this
Rafaela Gaspar ID 192763 BTEC Applied Science Level 3
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