The sample of water contains
#Ca^"++",NO_3^-,HCO_3^"-"and Cl^"-" " ions"# . We are to find out the temporary hardness of water.
The temporary hardness in this sample will be caused by #Ca(HCO_3)_2# which reacts with acid solution used for titration as follows.
#Ca(HCO_3)_2+H_2SO_4->CaSO_4+2H_2O+2CO_2#
This reaction reveals that 1 mol of #H_2SO_4# is required for nutralisation of 1 mol
#Ca(HCO_3)_2#which is equivalent to 1 mol #CaCO_3#
Now the titration result is that
#5cm^3 or 5xx10^-3 dm^3
0.01moldm^-3#
acid nutralises #25cm^3 or 25xx10^-3dm^3 #water sample.
So the no of moles of acid required
#="volume of acid"xx"its strength"#
#=5*10^-3dm^3xx0.01moldm^-3#
#=5xx10^-5mol#
As per above noted equation the amount of #Ca(HCO_3)_2# in water sample will also be
#=5xx10^-5mol#
#"In "CaCO_3" equivalent the no.of moles of "CaCO_3" will be"#
#=5xx10^-5mol#
Now molar mass of #CaCO_3# is
#=(40+12+3*16)g/"mol"=100g/"mol"#
So the mass of #CaCO_3#
#=5xx10^-5molxx100g/"mol"#
#=5xx10^-3g=5mg#
Now
#25xx10^-3dm^3# water sample contains 5 mg #CaCO_3# equivalent #Ca(HCO_3)_2#
So temporary hardness of water is
#=("Mass expressed in " CaCO_3" equivalent")/"Volume of water"#
#=5/(25xx10^-3)mgdm^-3#
#=200mgdm^-3#
A Short Cut
#"Hardness"=(SxxVxx50xx1000)/V_wmgdm^-3#
Where
#S="Strength of acid in
Normality"#
#"Normality"="Molarity"(M)xx"Basicity"(b)#
#V="Volume of acid in L"ordm^3#
#V_w="Volume of water sample in L"ordm^3#
