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.
