Question

A lake has been acidified to a pH of 4.210. If the lake has a volume of 5.32×104 cubic meters, what mass, in kg, of CaO must be added to raise the pH to a value of 7.00? Hint: The reaction of CaO with water is: CaO + H2O Ca(OH)2

Answer 1

the ph = 4.210 Volume af labe a 5.32x10" m² = 5.32 x 10 L final ph required 7 . Rxn GO + H₂O - Ca (012) Initial ph = 4.210 Phi - lag tt - 4.210 = - log Ht = t = 6.1X10SM H+ Concentration af af leke X € Mole af Ht a volume af leke x concente 5:32 x 107 x 6.1 xlos Initial = 3245.2 moles - Final Phar final concentection of uit = 101 Ph= 7 Ph= -long (+] 7 = -log (wit] st=107m mole afnt = volume of lake x conc. of tt = 5.328 107 x 107 moles it left = 5.32 moles of st moles of nt 1. Initial - moles af ht it reacted = moles of = 3245.2 - 5:32 = 3239.8 moles Can), Q2+ + 2014- 3239.8 mole af got react with 3239.8 mole of on- which is produced og half mole of on. Ca(OH)2 mole = 32S 9.8 = 1619.9 mole

As we kraw from this repction Сло + H, о Саои ). Речиця ЩО Aare moga oo o oo и) 2 So weight of GO required mole af Cao x mol Cao = 1619 9 x S6:07 amo 9 сао 8 21.3 8 - 90 • 821 КЗ