The hardness of `10^5` L of a sample of `H_2O` was completely removed by passing through a zeolite softener. The bed on exhaustion required 500 L of NaCl solution containing `15gL^(-1)` of `NaCl` for regeneration. Calculate the hardness of the sample of water.

Explanation

" Eq of "NaCl`-=Eq` of exhausted resin
`--" Eq of "Ca^(2+) and Mg^(2+)`
`-=Eq` of permanent hardness
`M of NaCl=(Wxx1000)/(Mwxx1000)=(15)/(58.5)`
Moles of " Eq of "`NaCl-=` moles or " Eq of "permanent hardness `M_1V_1=M_2V_2`
`(15)/(58.5)xx500L-=M_2xx10^(5)L`
`M_2(Ca^(2+) or Mg^(2+) `ion in hard water)
`=(15xx500)/(58.5xx10^(5))EqL^(-1)`
`=128.205xx10^(-5)`
`=50xx128.205xx10^(-5)gCaCO_3L^(-2)`
`=(50xx128.205xx10^(-5)xx10^(6)mL)/(1000mL)`
`=64.10 ppm of CaCO_3`
Hence, `pH=64.10`
Alternatively,
Let `V_1` and `V_2` be the volumes of hard water and `NaCl` solution containing `x g L^(-1)` of zeolite. Then, hardness (H) is given by
`H=(50xx x xxV_2xx10^(3))/(58.5xxV_1)` `{:[(Ew(CaCO_3)=50),(Mw(NaCl)=58.5)]:}`
`=(50xx15xx500xx10^(3))/(58.5xx10^(5))=64.10ppm`