Purpose
Determination of the concentration of hydrochloric acid in an unknown solution by setting opposite a urtiter substance (primary standard).
Method
A known amount of Tris, a known concentration titrated with the unknown solution of HCl.
Theory
One mole of hydrochloric acid reacts with a mole Tris. Analyses with hydrochloric acid has a molaritet between 0.08 and 0.12 M are examined, we calculated the approximate folder point to be at 16.5 ml added HCl (from the information that the relationship between drug volume of Tris and HCl is 1:1 and hydrochloric acid concentration on average is 0.1 M).
Reaction Equations
Fabric Equation:
(HOCH2) 3C-NH2 (aq) + HCl (aq) ® (HOCH2) 3C-NH 3 + (aq) + Cl-(aq)
Ionligning:
R-NH2 (aq) + H + (aq) ® R-NH3 + (aq)
Results
We chose the unknown acid # 4
Grovtitrering:
- Mass of Tris: 1.5448 g - 1.3251 g (tray weight + residue) = 0.2197 g
- We titrated 15.5 ml HCl
1st accurate titration:
- Mass of Tris: 1.5109 g - 1.3251 g (tray weight + residue) = 0.1858 g
- We titrated 12.8 ml HCl
2nd accurate titration:
- Mass of Tris: 1.5615 g - 1.3197 g (tray weight + residue) = 0.2418 g
- We titrated 16.9 ml HCl
Calculations
We calculate the concentration of the unknown acid solution:
Grovtitrering:
We calculate the first n (Tris):
As mentioned earlier reacting hydrochloric acid and tris 1:1, so
n (Tris) = n (HCl) = 0.0018 mol
So the concentration of hydrochloric acid is then:
1st accurate titration:
We calculate the first n (Tris):
So the concentration of hydrochloric acid is then:
2nd accurate titration:
We calculate the first n (Tris):
So the concentration of hydrochloric acid is then:
Average of the three provisions:
Discussion
Our tests were very successful. Grovtitreringen indicated that we had a hydrochloric acid concentration of 0.12 M. And after making two accurate titrations, we were unsubstantiated assumption, and we conclude therefore that our unknown acid (IV), the concentration 0.12 M.
Exercise 3b: Synthesis of iron (II) sulfate heptahydrate
Purpose
Production of iron (II) sulfate heptahydrate FeSO4 × 7H2O.
Method
Iron nails are dissolved in sulfuric acid under hydrogen development and formation of iron (II) sulfate in solution.
Theory
Iron is less electro-negative than hydrogen, and therefore will be formed iron (II) ions in the solution with sulfuric acid. We use an excess of iron, which causes precipitation of metals which are more electronegative than iron, and prevents oxidation of Fe2 + to Fe3 +.
Reaction Equations
Ionligning:
Fe (s) + 2 H + (aq) → Fe2 + (aq) + H2 (g)
Fabric Equation:
Fe (s) + H2SO4 (aq) + 7 H2O (l) → FeSO4 × 7 H2O (s) + H2; (g)
The following reaction will proceed by oxidation with atmospheric oxygen:
2 H2O (l) + 4 Fe2 + (aq) + O2 (g) + 8 OH-(aq) → 4 Fe (OH) 3 (s)
Results
The mass of jernsøm: 4.95 g
Quantity added sulfuric acid: 4 ml
Yield of iron (II) sulfate heptahydrate: 12.88 g
Calculations
Fe (s)
+
H2SO4 (aq)
+
7 H2O (l)
→
FeSO4 × 7 H2O (s)
+
H2 (g)
m
4.95 g
20.02 g
M
55.85 g / mol
278.022 g / mol
n
0.0886 mol
0.072 mol
0072
V
4 ml
c
18 M
As you can see, sulfuric acid is the limiting factor.
The theoretical yield is: 20.02 g.
Dividend rate:
Discussion
As seen, we've got a relatively small yield. This is mainly due to the reaction failed to run to completion, because when we stopped cooking there were still large hydrogen development around the seams, and we believe that if we eg. gave the half to full hour more, then there was formed a lot more FeSO4 × 7H2O. Another small source of error is that during the trial (especially filtration) may lose a little of the substance.
Exercise 3c: In vitro experiments
1st attempt
- 4 M NH4Cl:
pH read about. 5th
Reaction Equations:
NH4Cl (aq) → NH4 + (aq) + Cl-(aq)
NH4 + (aq) + H2O (aq) ⇌ NH3 (aq) + H3O + (aq)
pH is calculated to:
As seen, there is no large deviation between measured and calculated pH value.
- 2 M NH3:
pH read about. 11th
Reaction Equation:
NH3 (aq) + H2O (aq) ⇌ NH4 + (aq) + OH-(aq)
pH is calculated to:
pH values are very close, considered the highest we could measure with the indicator paper was the 11th
2nd attempt
We saw no color change or precipitate, but there was a gas production, and we measured the pH of the gas to 11
Ionligning:
NH4 + (aq) + OH-(aq) → NH3 (g) + H2O (l)
Fabric Equation:
NH4Cl (aq) + NaOH (aq) → NH3 (g) + H2O (l) + NaCl (aq)
When we measure the gas has a pH of approx. 11, it fits quite well with the ammonia in gaseous form is being formed.
3rd attempt
Change Color: clear blue → Blue (cloudy / cloudy) → dark blue (clear)
Precipitate: a drop of NH3 resulting dark blue precipitate, and 6 drops dissolve the precipitate.
Reaction:
When copper (II) sulfate dissolved. Kobberionen is (bright) blue in water:
CuSO4 (s) ® CU2 + (aq) + SO42-(aq)
Ammonia is a weak base:
NH3 (aq) + H2O (l) ⇌ NH4 + (aq) + OH-(aq)
Addition of ammonia to copper (II) sulfate gives first a precipitate of blue copper (II) hydroxide:
CU2 + (aq) + 2 OH-(aq) ⇌ Cu (OH) 2 (s) (*)
In excess of ammonia formed by deep mørkebå kompleksion tetraamminkobber (II) ion:
CU2 + (aq) + 4 NH3 (aq) ⇌ Cu (NH3) 42 + (aq)
Doing shift equation (*) to the left, copper (II) hydroxide is dissolved, and obtained a very dark blue solution:
Cu (OH) 2 + 4 NH3 (aq) ⇌ Cu (NH3) 42 + + 2 OH
4th attempt
Change Color: clear milky → → clear
Precipitate: a drop of NH3 resulting white precipitate, and 7 drops leads to precipitate dissolved.
When zinc sulphate is dissolved:
ZnSO4 (s) ® Zn2 + (aq) + SO42-(aq)
Addition of ammonia to zinc sulphate gives first a precipitate of white zinc hydroxide:
Zn2 + (aq) + 2 OH-(aq) ⇌ Zn (OH) 2 (s) (*)
In excess of ammonia formed the complex ion tetraamminzink (II) ion:
Zn2 + (aq) + 4 NH3 (aq) ⇌ Zn (NH3) 42 + (aq)
Doing shift equation (*) left, zinc hydroxide is dissolved, and obtained a clear solution:
Zn (OH) 2 (s) + 4 NH3 (aq) ⇌ Zn (NH3) 42 + + 2 OH-(aq)
