Question

Assuming complete dissociation, what is the pH of a 4.67 mg/L Ba(OH)2 solution?

Answer 1

we know that

concentration in mol / L = (conc in g / L) / molar mass

also

molar mass of Ba(OH)2 = 171.34

given

conc in g / L = 4.67 x 10-3 g / L

so

conc in mol / L = 4.67 x 10-3 / 171.34

conc in mol / L = 2.7255 x 10-5

now

Ba(OH)2 ---> Ba+2 + 2OH-

we can see that

[OH-] = 2 x [Ba(OH)2]

so

[OH-] = 2 x 2.7255 x 10-5

[OH-] = 5.45 x 10-5

now

pOH = -log [OH-]

so

pOH = -log 5.45 x 10-5

pOH = 4.264

now

we know that

pH = 14 - pOH

so

pH = 14 - 4.264

pH = 9.736

so

the pH of the solution is 9.736