Question

Calculate carbonate and noncarbonate hardness of water sample containing following dissolved salts in mg/L: Mg(HCO3)2 = 58.5 , Ca(HCO3)2 = 40.5, MgSO4 = 3, Ca(NO3)2= 41, CaCl2 = 33.3, SiO, = 10 MgCl, = 9.5​

Answer 1

Ques

Calculate the temporary and permanent hardness of water sample having the following constitute per liter

Ca(HCO3)2=162mg

MgCl2=95=mg

NaCl2=585mg

Mg(HCo3)2=73mg

CaSO4=136mg

Ans

Mole of Ca(HCO3)2=162×10-³mg÷162g/mole =1*10-³mole.

Mole of Ca(SO4) =136*10-³g÷136g/mole=1*10-³mole.

Total Mole of Ca =2*10-³mole.

mass of CaCo3=2×10-³×100

=0.2g

therefore, ppm(permanent hardness)=6.2/1000×10⁶=200ppm.

Mole of MgCl2=95×10-³/95=1×10-³mole.

Mole Mg(HCg)2=73*10-³÷146=5×10-⁴mole.

Mole Mg=1.5×10-⁴mole.

Mole g CaCo3(in terms of mg) =1.5×10-³

mass=1.5×10-³=0.150g.

ppm(temporary hardness)=0.150/100*106.

=150ppm.

this is answer.