In this example problem, we will write the molecular, complete ionic, and net ionic equations for three different acid-base reactions. Our problem reads: write the balanced molecular, complete ionic, and net ionic equations for the reactions between: 1. Sodium hydroxide and hydroiodic acid 2. Sodium hydroxide and hydrofluoric acid 3. Sodium hydroxide and sulfuric acid First, we'll write the molecular equation between sodium hydroxide and hydroiodic acid. This is a reaction between a strong base and a strong acid. When we write our reaction, sodium hydroxide reacts with hydroiodic acid to produce water and the salt sodium iodide. This equation is already balanced with just ones as coefficients. Next, we'll write the complete ionic equation for this reaction by splitting apart strong electrolytes. Both sodium hydroxide and hydroiodic acid are strong electrolytes. Hydroiodic acid is a strong acid, so we'll split apart sodium hydroxide, hydroiodic acid, and the soluble salt sodium iodide. We do not split apart H2O because it is not a strong electrolyte. To write our net ionic equation, we're going to eliminate the spectator ions from the complete ionic equation. The sodium ion shows up on both sides of the equation, and the iodide ion shows up on both sides of the equation. So our net ionic equation is: hydroxide + H+ produces water. These are our molecular, complete ionic, and net ionic equations for the reaction between sodium hydroxide and hydroiodic acid. Next, we'll write the molecular equation between sodium hydroxide and hydrofluoric acid. This is a reaction between a strong base and a weak acid. Sodium hydroxide and hydrofluoric acid combine to produce water and sodium fluoride. To write the complete ionic equation, we need to split apart strong electrolytes. Here, the only strong electrolyte is sodium hydroxide and sodium fluoride, which are soluble salts. Hydrofluoric acid is a weak acid, so it is not a strong electrolyte. So we have to keep hydrofluoric acid together as aqueous HF. To write the net ionic equation for this reaction, from the complete ionic equation, we just need to eliminate the spectator ion. And what's left is: hydroxide + hydrofluoric acid produces water + fluoride. These are our molecular, complete ionic, and net ionic equations for the reaction between sodium hydroxide and hydrofluoric acid. Next, we'll write the molecular equation for the reaction between sodium hydroxide and sulfuric acid. This time, we need two sodium hydroxide to react with sulfuric acid because sulfuric acid is a diprotic acid and it will produce two waters and sodium sulfate. To write the complete ionic equation, we'll have to break apart our strong electrolytes. Sodium hydroxide is a strong electrolyte, so we break it apart to produce two sodium+ and two hydroxide ions. H2SO4 (sulfuric acid) is a diprotic acid, and only the first hydrogen is a strong acid. The second hydrogen is a weak acid. So we split this apart into H+ from the strong acid and HSO4from the weak acid. And we will produce two water + two Na+ and sulfate. We split apart sodium sulfate because it's a soluble salt and a strong electrolyte. Our last step is to write the net ionic equation by eliminating our spectator ions. Our only spectator ions are our two sodium+ ions. So everything that's left is: two hydroxide + H+ + hydrogen sulfate produces two water + sulfate. These are our molecular, complete ionic, and net ionic equations for the reaction between sodium hydroxide and sulfuric acid.