Strong acids and caustic alkalis can cause chemical burns. Although one can kill the other, in case of an accident you should not use the other, as the resulting reaction would be too vigorous and exothermic ( heat producing). The best thing would be to softly flush the affected area with tap water for about 10 minutes. |
Testing concentrations.
Titration is the method used to determine the strength of
a solution using a solution of known strength. The apparatus used for this
purpose is depicted above.
Very often this is employed for acid base reactions. If
you wish to determine the strength of a sodium hydroxide solution, you can take
a measured olume into a titration flask. The burette can be filled with the
acid of known strength. We have to add an indicator to find the exact point of
neutralization.
Acid and Base
reactions
Consider this reaction: NaOH +
HCl =
NaCl + H2O
Mole ratio:- ………… 1mol :
1mol
This shows that 1mol of acid can completely neutralise 1 mol of the
base.
This shows that 1.L of
1M-NaOH can be completely neutralised by 1L of 1M – HCl.
When the reacting mole ratio is 1:1,
equal volumes of identical strength can neutralize each
other.
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Then the formula is:-
MaVa Product of molarity and volume of acid.
MbVb Product of molarity and volume of base
Example 1.
25 ml of a 0.1 M- sodium hydroxide solution was completely neutralised by 21.2 ml of HCl. Find the strength of the acid.
Equation:- NaOH + HCl ------> NaCl + H2O.
Molar ratio:- 1 : 1
M a x V a = M b x Vb
M a x 21.2 = 0.1 x 25
M a … = 0.1 x25/ 21.2 = 0.12 M
We may consider this reaction in the ionic form.
What actually happens is this:-
NaOH <-------> ionises as ( Na+ ) + ( OH-)
HCl <-------> ..................( H+ ) + (Cl-)
While sodium ions and chloride ions remain in the solution as they are hydrogen ions combine with hydroxyl ions forming water molecules.
So simply the reaction is :-
H+ + OH- -------> H2O
Ratio : 1 : 1
When the concentrations of the two solutions are not the
same we cannot use this formula:-
In that case the formula has to be :-
Ra: Rb is the neutralizing ratio of acid and base.
Example 2
2NaOH + H2SO4 = Na2SO4 + 2H2O
Mole ratio: 2 :
1.
Sodium hydroxide volume - 1000
ml strength 2M
Sulphuric acid volume Va
strength 3M
Find the volume of sulphuric acid required.
Answer
No. of moles in 1.L of NaOH……2
H2SO4 moles required for 2 mol. of NaOH ..1
Volume of H2SO4 that
has 3 mol. 1.L
Volume of H2SO4
……………1 mol =
1/3 L
= 333.33 mL.
Answer using Formula
M a x V a =
M b x Vb
EMa
EMb
3 x V = 2
x 1000
1 2
V =
2000/6 = 333.33 mL.
Example
3
Sodium
carbonate = Mass ….- 5.3g. Volume 250 mL.
HCl
required to neutralise = Volume 30 mL.
Find
the strength of the acid solution.
Answer
Equation:- Na2CO3 +2 HCl ----à 2 NaCl + H2CO3
First of
all we have to find the molarity of the base.
Mass
dissolved in 250 mL = 5.3g
Mass……………1000 mL = 5.3x4 = 21.2 g
Converting
this to moles = 21.2/106 = 0.2moles
Formula M
a x V a =
M b x Vb
EMa EMb
Substitution: M
x 30 = 0.2 x 25
Mole ratio 2 : 1
M
= 0..2x25x2/ 30 =0.333
Eample
4
2M
Hydrochloric acid ,100 mL, was added to 10g of
mortar containing lime and sand
Assume that CaO was the only reactant present . The remaining solution
was filtered and titrated with a 2M –NaOH solution. The volume of alkali
required was 75.0 mL. Find the percentage of CaO in mortar.
First equation. CaO + 2 HCl ----à CaCl2 + H2O
Second equation NaOH
+ HCl
----à NaCl + H2O
We have to work on the second equation first.
As the mole ratio is 1:1 we can use
M a x V a =
M b x Vb
2 x V =
2 x 75
V = 75 mL of acid,
Volume of acid added to mortar
=100 mL
Volume of acid reacted = 100-75 = 25 mL
No. of moles of acid in 25 mL 2x25/1000
= 50/1000 = 1/20
=0.05
CaO +
2 HCl ----à CaCl2 + H2O
Mole
ratio 1 :
2.
2 mol
of acid can react with 1 mol of base,
0.05…………………………..1/2 x0.05 = 0. 0 25 mol.
Mass of 1 mole of CaO = 56 g
0.025mol =
56 x0.025 = 1.4g
1.4g present in 10 g of mortar.
Mass
of CaO in 10g of mortar ….= 1.4 g
Mass…………………….100g……= 14g.
As
a Percentage of CaO = 14 %. Of CaO.
1.0
Write the equation for each reacti0n and determine what is not given.
Acid strength | Acid Volume | Base strength | Base volume |
1…. HCl + NaOH ------> NaCl + H2O |
1 M- HCl | 25 mL | 1M =NaOH | i.…………. |
2. |
2M - HCl | 25 mL | 1 M -NaOH | ii…………... |
3. H2SO4 + 2 NaOH -------> Na2SO4 + 2 H2O |
1M= H2SO4 | 25 mL | 1 M - NaOH | iii………………... |
4. 2M=H2SO4
| 25 mL | 5 M- NaOH | iv……………. |
5. HNO3 + KOH ------> KNO3 + H2O |
2M-HNO3 | 10 mL | V,,,,,, KOH | 50 mL |
Answers
I. 25 mL. Ii. 50 mL. Iii. 50 mL. Iv. 20 mL v, 0.4 M.
2.0 Find the answers for these.
i. Give the reacting molar ratios for the reactants and products for this equation:
2 HCl + K2CO3 ------> 2 KCl + H20 + CO2
ii. What is the volume of 1 M- HCl required to react with 0.5 mol of K2CO3
iii. If 0.05 L of potassium carbonate produced a residue of 5.8g on evaporation what would have been the molarityof the solution?
iv. If 18.5 mL of HCl could neutralise 25 mL of 0.5 M potassium carbonate solution, what is the strength of the acid?
v. How many moles of CO2 will be produced by the above reaction?
ANSWERS
i. 2: 1 ------> 2 : 1:1
ii. To react with 1 mol. of K2CO3 volume of acid required = 2 mol.
0.5 mol...................................................... = 1 mol.
iii.
Mass of solute in 0.05 L of solution = 5.8g
Mass ................... 1 L .....................= 5.8/ 0.05 = 116g.
Converting 116 g to moles .............=116/138.2 = 0.84 mol.
Molarity of solution ................ = 0.84 M.
3.0
Fill in the blank spaces in this table so as to neutralise each other.
Equation:- KOH + HCl
-----à KCl
+ H2O
Acid Volume
|
Acid strength
|
Base volume
|
Base strength
|
20mL
|
1-M.
|
10 mL
|
i.
…………….
|
12 mL
|
2-M
|
ii ……
|
3-M
|
15 .5 mL
|
iii…………
|
20 mL
|
0.5 M.
|
iv…………
|
0.1 M
|
0.25 L
|
0.1 M
|
25mL
|
v…………..
|
22 mL
|
5.6g/L
|
4.0 This question is about acetic acid in vinegar.

i. What is the molar mass of this acid.
ii. A sample of home vinegar has 4% acetic acid. Assuming 100g
of water to have 4g of vinegar find the molarity of the sample..
iii. How many grams of CO2 can you obtain by the
reaction of 25mL of this vinegar on baking powder.
Word
equation:-
Sodium bi
carbonate + acetic acid -------à sodium acetate + water + carbon dioxide.
Equation:- ……………….NaHCO3 + CH3COOH → CH3COONa + H2O + CO2
Answers
i. CH3COOH =
12+3+12+32+1 = 60g/mol
ii.: As the
density of water is 1g /cm3, the volume of 100g of water = 100ml.
100mL will have 4g of acid.
1000 mL……………………..40g.
Converting this to moles
,,, = 40/60
= 2/3 = 0.67 M
iii.
Equation: NaHCO3 + CH3COOH
→ CH3COONa
+ H2O + CO2
Molar
ratio: 1 : 1 1
1000 ml acid solution
has 0.67 mol
25 mL………………………..= 0,67 x 25/1000
=0.017 mol
CO2 produced by this will be same number of moles = 0.017 mol
1 mol of CO2 ……………………….44g
0.017 mol………………………….= 44x0.017 = 0.75g.
Write the equation for each reaction and determine what is not given.
Acid strength | Acid Volume | Base strength | Base volume |
1…. HCl + NaOH ------> NaCl + H2O |
1 M- HCl | 25 mL | 2M =NaOH | i.…………. |
2. |
2M - HCl | 25 mL | 1 M -KOH | ii…………... |
3. H2SO4 + 2 NH4OH -------> (NH4)2SO4 + 2 H2O |
1M= H2SO4 | 25 mL | 1 M - NH4OH | iii………………... |
4. 3M=H2SO4
| 25 mL | 5 M- NaOH | iv……………. |
5. HNO3 + KOH ------> KNO3 + H2O |
0.5 -M-HNO3 | 10 mL | V,,,,,, KOH | 50 mL |
1.0 This question is about sodium carbonate and hydrochloric acid.
Equation:- Na2CO3 + 2 HCl -------> 2 NaCl + H20
A decimolar solution (0.1-M) of sodium carbonate was titrated to find the strength of hydrochloric acid. 0.25 L of Na2CO3 required 0.20.1 L of acid.
Find the following:
i. Molar massof sodium carbonate.
ii. No. of moles in 0.25 L. of sodium carbonate solution.
iii. No of moles of HCl that the abobe 0.25 L can neutralize.
iv. The strength of the acid solution.
2.0 This question is about
magnesium hydroxide. Milk of magnesia ,Mg(OH)2 is a mild alkali that
was used to neutralize excessive hydrochloric acid in the stomach. i.
What is the molar
mass of Mg(OH)2 /
ii. This substance was used to
reduce acidity in the stomach. Acid molarity in digestive juice can go up to
o.1M. of HCl. Find the mass of Mg(OH)2 that can neutralize acids in 20 mL of juice
(5x2=10 points)
3.0
25 mL of Phosphoric acid was used to react with 7.4
g of calcium hydroxide .
Equation:
3 Ca(OH)2
+2 H3PO4------à Ca3(PO4)2 + 6 H2O
i. Molar mass of the
acid.
ii.
No. of Ca(OH)2 moles in 7.4 g.
iii.
Molarity of acid
iv.
Mass of calcium phosphate formed.
(5x4= 20)