Question

A water treatment plant treating 10 MLD of water requires 20 $mg/L$ of filter Alum $A{l}_2(S{O}_4{)}_3.18H_{2}0$. Water has 6 $mg/L$ of alkalinity as $CaC{O}_3$ .
( Al = 26.97, S = 32, 0 = 16, H = 1, Ca = 40, and C = 12)
Total alkalinity requirement (${10}^6$ mg per day as $CaC{O}_3$) matching filter alum, shall be

• A. $120$
• B. $60$
• C. $90$
• D. $180$

Answer 1

The correct option is C $90$

In water treatment, water must have sufficient alkalinity for alum to be effective.
Reaction of alum with $HC{O}_3^{-}$ alkalinity is

$\underset{\left(666g\right)}{A{l}_2(S{O}_4{)}_3}.18H_{2}O+\underset{(3.\times 162g)}{3Ca(HC{O}_3{)}_2}\to$

$3CaS{O}_4+2Al(OH{)}_3\downarrow +6C{O}_2+\uparrow 18H_{2}O$

Discharge, $Q=10MLD$

$=10\times {10}^{6}L/d$

Alum required per day

$=10\times {10}^6\times 20mg/d$

$=200\times {10}^{6}mg/d$

666 g alum require alkalinity

$=3\times 162g$

$200\times {10}^{6}mg$ alum require alkalinity

$=\frac{3\times 162}{666}\times 200\times {10}^6$

$=145.95\times {10}^{6}mg/d$

Alkalinity expressed as

$CaC{O}_3=145.95\times {10}^6$

$\times \frac{\text{Equivalent weight of}CaC{O}_3}{\text{Equivalent weight of}Ca(HC{O}_3{)}_2}$

=$145.95\times {10}^6\times \frac{50}{81}$

$=90.09\times {10}^{6}mg/d$

as $CaC{O}_3=90({10}^6$ mg per day as $CaC{O}_3)$