Question

Calculate the pH of 0.03M Ba(OH)₂.

a. 10.78

b. 12.78

c. 2.78

d. 1.22

e. 11.78

Answer 1

Answer:- option b. pH = 12.78

here it is given that

[Ba(OH)2] = 0.03 M

hence we know that Ba(OH)2 will dissociates into ions when we add in water

so the dissociation reaction will be

Ba(OH)2 $\rightarrow$ Ba2+ + 2OH-

0.03 0.03 0.06

hence we know that

pOH = - log [OH-]

pH = 14 - pOH

pH = 14 + log[0.06]

pH = 12.778

or pH = 12.78