Question

Calculate the amount of lime (85% pure) and soda (95% pure) required to soften one million litre of water which contains CaCO₃ = 12. 5ppml MgCO₃ = 8.4ppml
CaCl₂ = 22.2ppml ,  MgCl₂ = 9.5ppml C0₂ = 33ppml HCI = Z3ppml  organic matter= 16.Bppm 

Answer 1

lmpurities(mg/lit)	Multiplication factor	CaC03 equivalent (mg/lit)	Requirement
CaC03=72.5	`100/100`	`12.5xx 100/100 = 12.5 `	L
MgC03=8.4	`100/84`	`8.4 xx 100/84= 10`	2L
CaCl2 = 22.2	`100/111`	`22.2 xx 100/111 = 20`	S
MgCl2 = 9.5	`100/95`	`9.5 xx 100/95 = 10`	L+S
C02 = 33	`100/44`	`33 xx 100/44=75`	L
HCl = 7.3	`100/73`	`7. xx 100/73=10`	L+S

Nacl  does not react with lime and soda.
`LIME=74/100 [caC0_3  "equivalent of " 2 xx MgC0_3 + CaC0_3 + MgCl_2 + HCl+CO_2 ] xx`

`("volume of water")/ 100 xx 100/ (%"purity")`

`74 /100 =xx [2 xx 10 + 12.5 + 10 + 10 + 75] xx 10^6/1000 xx 100/ 85`

= 111000 gms.

`SODA =106/100  [caCO_3 " equivalent of " CaCl_2 + MgCl_2 ] xx ("Volume of water")/ 1000 xx 100/("%purity").`

`= 106/100  [20 + 10] xx 10^6/1000 xx 100/95.`

= 33473.68 qms.

The lime requirement is 111000 gms and soda requirement is 33473.68 gms.