Question

A waste water that contains sulfuric acid is found to have a concentration of 0.105 Normality. This will be subjected for neutralization. If the estimated quantity of the waste water is 1000 liters per day, how many kilograms of calcium hydroxide will be required to neutralize the waste water in a day?

Answer 1

The equation for the reaction is;

$Ca(OH)_2+H_2SO_4\to CaSO_4+H_2O$

Total volume$=1000L$

Concentration of $H_2SO_4=0.105m/L$

$\therefore$ $1000LH_2SO_4\times$ $0.105moles\over 1L$ $H_2SO_4$ $=105 molesH_2SO_4$

Mole ratio of $H_2SO_4$ and $Ca(OH)_2$

$=1:1$

$1mole H_2SO4\times$ $0.105moles Ca(OH)_2\over105moles H_2SO_4$ $=0.001moles CaOH$

$0.001moles Ca(OH)_2\times$ $74.093g/mol Ca(OH)_2\over1mol Ca(OH)_2$ $=0.074gCa(OH)_2$

$=7.4\times 10^{-5}Kg$