Let
V_(HCl)mL->"Volume of HCl solutoin"
S_(HCl)M->"Strength of HCl solutoin"
V_(Al(OH)_3)mL-> "Volume of "Al(OH)_3" solutoin"
Amount HCl taken=V_(HCl)*S_(HCl)mmol
Amount of Al(OH)_3 taken
=V_(Al(OH)_3)xx0.22" "mmol
The reaction
Al(OH)_3 +3HCl->AlCl_3+3H_2O
This equation reveals that the reacting ratio of no. of moles is
Al(OH)_3):HCl= 1:3
So the residual amount of Al(OH)_3 is
=(V_(Al(OH)_3) xx 0.22- 1/3xxV_(HCl)*S_(HCl))mmol
The amount of H_2SO_4 required for neutralisation of ecess Al(OH)_3 is 5mL 0.38 M
equiv5xx0.38mmol
Equation of neutralisation of excess Al(OH)_3 by H_2SO_4
2Al(OH)_3 +3H_2SO_4->Al_2(SO_4)_3+6H_2O
This equation reveals that molar reacting ratio is
Al (OH)_3:H_2SO_4=2:3
So
V_(Al(OH)_3) xx 0.22- 1/3xxV_(HCl)*S_(HCl) =2/3xx5xx0.38
Using this equation we can find out S_(HCl) if the value of V_(HCl) of and V_(Al(OH)_3) are known.
