Question

An engineer added 100 lb of CaO (70% pure) to the tank, and then he found the pH was too high. How many lb of HCl (39% pure) is needed to readjust the pH to 7.5?

Answer 1

CaO + H2O = Ca(OH)2

Ca(OH)2 --> Ca2+ + 2OH-

then

1 mol of CaO = 2 mol of OH-

mass of CaO = 100*0.70= 70 lb of CaO

mass in g = 70 lb * 454 g/lb = 31780 g of CaO

mol = mass/MW = 31780 / 560774 = 0.05667 mol of CaO

that is,

0.05667 mol ofCaO = 2*0.05667 = 0.11334 mol of OH-

if pH = 7.5 required

pOH = 14-pH = !4-7.5 = 6.5

[OH-] --> 10^-pOH

[OH-] = 10^-6.5 = 3.162*10^-7 M

0.11334 mol of OH- - X mol of HCl --> 3.162*10^-7mol of OH- left

X = 0.11334 - 3.162*10^-7

X = 0.1133396838 mol of H+ required

mol of HCl --> 1:1 ratio

mol of HCl = 0.1133396838 mol requires

mass = mol*MW = 0.1133396838*36.4609 = 4.13246687706 g

mass in lb --> 4.13246687706/454 = 0.009102 lb

for 39% purity --> 0.009102/0.39 = 0.02333 lb of HCl must be added