Convert mol/given vol.(acid/base) to conc(acid/base). in mol/dm3
This method is called FIRST PRINCIPLE
Slide 42
The tips in chart
acid acid
Molar conc. Conc. in given vol. mole ratio Conc. In given vol. molar conc.
base base
Mass conc.
Mass conc.
Slide 43
Examples
20cm3 of tetraoxosulphate (vi) acid was neutralized with 25cm3 of 0.1mold-3 sodium hydroxide solution. The equation of reaction is
H2SO4 + 2NaOH Na2SO4 + 2H2O
Calculate (i) conc. of acid in moldm-3 (ii) mass conc. of the acid.
[H=1, S= 32, O=16]
Slide 44
Given:
conc. of the base = 0.1moldm-3
Vol. of the base = 25cm3
Convert to conc. in given vol.
0.1 mol in 1000cm3
X mol in 25cm3
X = 0.1 x 25
1000
0.0025mol(per25cm3)
Use mole ratio
Acid : base
1 : 2
X : 0.0025
X = 0.00125mol(in given vol of the acid) i.e 20cm3
Convert to conc.(acid) in moldm-3
0.00125mol in 20cm3
X in 1000cm3
X = 0.0625mol.
.: conc. of acid
= 0.0625moldm-3
Slide 45
Example 1 continues
ii mass conc. of the acid :
Mass conc. = molar conc. X molar mass
0.0625 x [2+32+64]
0.0625 x 98=6.13gdm-3
Remember, always leave your answers in 3 s.f.
Slide 46
More
If 18.50cm3 hydrochloric acid were neutralized by 25cm3 of potassium hydroxide solution containing 7gdm-3. what is the conc. of the acid in moldm-3?
The equation of reaction:
HCl + KOH HCl + H2O
[K = 39, O = 16, H = 1]
Slide 47
Given:
Mass conc. of the base = 7gdm-3
Convert to moldm-3 : Mass conc. = 7
Molar mass [39+16+1]
= 0.125 moldm-3
Mol reacted at the given vol.(25cm3)
n = conc. in moldm-3 x vol.(dm3) 0.125 x 25/1000
0.003125mol
Using mole ratio
Acid : base
1 : 1
X : 0.003125
X = 0.003125
0.003125mol[per18.5cm3] in moldm-3
0.003125mol in 18.5cm3
X in 1000cm3 x = 0.169mol
.: conc. of the acid
= 0.169 moldm-3
Slide 48
You can use “the theory”
CaVa = na
CbVb nb
Example 1 again.
20cm3 of tetraoxosulphate (vi) acid was neutralized with 25cm3 of 0.1mold-3 sodium hydroxide solution. The equation of reaction is
H2SO4 + 2NaOH Na2SO4 + 2H2O
Calculate (i) conc. of acid in moldm-3 (ii) mass conc. of the acid.