Answer;
[Ba2+] = 0.550 M
[OH-] = 1.10 M
Explanation;
Each mol Ba(OH)2 releases 2 mol OH-
Ba(OH)2 -----> Ba2+ + 2OH-
Mole ratio; Ba(OH)2 :Ba2+ = 1 : 1
[Ba2+] = 0.550 M
Mole ratio; Ba(OH)2: 2OH- = 1 : 2
= 2 × 0.550 = 1.10
[OH-] = 1.10 M
This assumes complete ionization (dissociation) of Ba(OH)2.
