Question

Find the Ph:

of 7.01 x 10-5 M Ba(OH)2

thanks:)

Answer 1

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

[Ba(OH)2] = 7.01E-5
[OH^-] = 2*7.01E-5
pOH = -log(OH^-)
pH + pOH = pKw = 14.