Question

Approximately how many pounds of calcium oxide, CaO, must be added to the water to achieve a pH of 3.88?

Answer 1

If we put CaO in water, that converts into Ca(OH)2 as shown below

Balanced equation:
CaO + H2O ====> Ca(OH)2
Reaction type: synthesis

Let us take one liter of water

pH of the solution required = 3.88

pH + pOH = 14

pOH = 14 - 3.88 = 10.12

pOH of the solution required = 10.12

Let us calculate the concentration of OH-

pOH = - Log OH-

10.12 = -LogOH-

OH- =  7.5857 x 10^-11 M

We have to prepare a solution with  7.5857 x 10^-11 M of OH- concentration

Ca(OH)2 concentration required = 3.7922 x 10^-11 M

Because each moles of Ca(OH)2 will provide 2 moles of OH-

CaO concentration required = 3.7922 x 10^-11 M

Molar mass of CaO =  56.0774 g/mol

Mass of CaO required in gram = 3.7922 x 10^-11 mol x 56.077 g/mol =   2.126 x 10^-9 g

Mass of CaO in pound = 4.689 x 10^-12 pound