Michael S. answered • 13d
B.S. in Chemistry, Indiana University; Organic Chem Teaching Intern
Net ionic equation:
OH−(aq) + H+(aq) → H2O(l)
That is the whole thing. Here is how you get there, because the work matters more than the answer on this type of problem.
Step 1 — molecular equation, balanced
Ba(OH)2(aq) + 2 HI(aq) → BaI2(aq) + 2 H2O(l)
The 2 in front of HI is required — barium hydroxide supplies two hydroxides per formula unit, so it takes two acid protons to neutralise it.
Step 2 — complete ionic equation
Break apart everything that is strong and aqueous. Ba(OH)2 is a strong base, HI is a strong acid, and BaI2 is soluble, so all three dissociate. Water does not.
Ba2+ + 2 OH− + 2 H+ + 2 I− → Ba2+ + 2 I− + 2 H2O
Step 3 — cancel the spectators
Ba2+ and I− appear unchanged on both sides, so both are spectators. Removing them:
2 OH− + 2 H+ → 2 H2O
Step 4 — reduce to lowest whole-number coefficients. Every coefficient divides by 2:
OH−(aq) + H+(aq) → H2O(l)
Three places students lose points here
1. Forgetting step 4. Leaving it as 2 OH− + 2 H+ → 2 H2O is marked wrong. A net ionic equation must be in lowest terms.
2. Writing BaI2 as a solid. Check solubility rules: iodides are soluble except with Ag+, Pb2+, and Hg22+. Barium is none of those, so BaI2 stays dissociated and cancels out. If it had precipitated, it would appear in the net equation and the answer would look completely different.
3. Not splitting HI. HI is one of the seven strong acids (HCl, HBr, HI, HNO3, H2SO4, HClO4, HClO3). If the acid had been weak — acetic acid, say — it stays intact as a molecule and the net ionic equation becomes CH3COOH + OH− → CH3COO− + H2O instead.
Worth memorising: every strong acid + strong base neutralisation with no precipitate gives the same net ionic equation, H+ + OH− → H2O. Once you recognise the pattern you can write it down immediately — but you still have to check the solubility of the salt first, because that check is what tells you the pattern applies.
