Question #0bff5

1 Answer
Jun 28, 2016

Answer:

#=200mgdm^-3#

Explanation:

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#