Question

What is the pH of a solution prepared by dissolving 5.00 g of Ca(OH)2 in 5.00 liters of water. Assume no volume change after Ca(OH)2 is added. (MM of Ca(OH)2 = 74.1 g/mol)

Answer 1

no. of moles = gm of compound / molar mass

molar mass of Ca(OH)2  = 74.1 g/mol

gm of Ca(OH)2 = 5.00

no.of moles of Ca(OH)2  = 5 / 74.1 = 0.0675 moles

molarity = no. of moles / volumes of solution in lliter

molarity of Ca(OH)2 = 0.0675 / 5 = 0.0135 M

Ca(OH)2 is strong base therefore dissociate completly in water as

Ca(OH)2  + H2O $\rightarrow$ Ca2+(aq) + 2OH-(aq)

According to dissociation reaction 1 mole Ca(OH)2 produce 2 moles 2OH-(aq) molar ratio of a(OH)2 to OH- is 1:2 therefore concentration of OH- is 0.0135 M X 2 = 0.0270 M

concentrarion of OH- = 0.0270 M

pOH = -log[OH-] = -log(0.0270) = 1.57

pH = 14 - pOH = 14 - 1.57 = 12.43

pH = 12.43