Strong acid–strong base reactions | Acids and bases | AP Chemistry | Khan Academy

Hydrochloric acid is an example of a strong acid, and sodium hydroxide is an example of a strong base. When an aqueous solution of hydrochloric acid reacts with an aqueous solution of sodium hydroxide, the products are an aqueous solution of sodium chloride and water. Since the reaction goes to completion instead of an equilibrium arrow, there's just an arrow going to the right, and this is called an acid-base neutralization reaction.

Let's think about the ions that are involved in this acid-base neutralization reaction. Since hydrochloric acid is a strong acid, HCl ionizes 100% in solution; therefore, in solution, HCl consists of H⁺ ions and Cl⁻ anions. Sodium hydroxide is a strong base, and strong bases dissociate 100% in solution; therefore, an aqueous solution of sodium hydroxide consists of sodium ions (Na⁺) and hydroxide anions (OH⁻).

For our product, sodium chloride is a soluble salt and, therefore, forms an aqueous solution. So, in solution, we would have sodium cations (Na⁺) and chloride anions (Cl⁻). Since water ionizes only to a very small extent, we don't write it as the ions; we simply write H₂O. To save some time, I went ahead and added in aqueous subscripts, and for water, it will be liquid, along with some plus signs, and the reaction arrow. This is called the overall ionic equation for our strong acid-strong base reaction.

Instead of calling it an overall ionic equation, we could also call it a complete ionic equation. We can use the overall ionic equation to determine the net ionic equation for this strong acid-strong base reaction. To figure out the net ionic equation, we first need to determine the spectator ions. Remember, spectator ions don't participate in the chemical reaction.

Notice how we have a sodium cation on the reactant side and a sodium cation on the product side, so we can cancel out the sodium cation. We also have a chloride anion on the reactant side and a chloride anion on the product side, so we can cross that ion out as well. So, the sodium cation and the chloride anion are the spectator ions. Once we take out our spectator ions, we're left with the net ionic equation.

So, for the net ionic equation, we would have H⁺ + OH⁻ → H₂O. This is one way of writing our net ionic equation. However, remember that H⁺ and H₃O⁺ are used interchangeably in chemistry. So, instead of writing H⁺ in our net ionic equation, we could have also written H₃O⁺. When H₃O⁺, or the hydronium ion, reacts with the hydroxide anion, we still form water, but notice for this version of the net ionic equation, we need to put a two as a coefficient to balance everything.

Next, let's think about a situation where we have equal moles of our strong acid and strong base. For example, let's say we have 1 mole of HCl and 1 mole of NaOH. To help us think about what's happening in this situation, we're going to use an ICF table, where I stands for the initial number of moles, C stands for the change in moles, and F stands for the final amount of moles. The reason why we're using an ICF table instead of an ICE table is that in an ICE table, the "E" stands for equilibrium, and here we assume the reaction goes to completion. Therefore, instead of "E," we're writing "F" for final amount of moles.

Let's fill out the rest of our ICF table. We have the initial amount of moles of HCl and NaOH. If we assume the reaction hasn't happened yet, the initial amount of moles of NaCl would be zero. Looking at the balanced equation, one mole of HCl reacts with one mole of sodium hydroxide. Since the reaction goes to completion, we can go ahead and write under the change part here we're going to react 1 mole of HCl, so we're going to lose that 1 mole of HCl.

Since it's a 1:1 ratio, we're going to lose 1 mole of NaOH. Since there's a one as a coefficient in front of NaCl, under the change part on our ICF table, we would write plus 1 mole for NaCl. Since we started with 1 mole of HCl and we lost one mole, the final amount of moles for HCl would be zero. Same thing for sodium hydroxide; the final amount of moles would be zero.

It's 1 minus 1 for NaCl. We started off with zero, and we gained one, so we're ending up with a final amount of moles of 1 mole of sodium chloride. When the reaction goes to completion, the strong acid and the strong base have completely neutralized each other, and essentially we have an aqueous solution of sodium chloride. At 25°C, the pH of water is 7, which is neutral, and the sodium cations and the chloride anions do not react with water, so they don't affect the pH. Therefore, for the situation where you have equal amount of moles of strong acid and strong base, the pH of the resulting solution will be seven.

Let's do another strong acid-strong base calculation. However, this time, the moles of strong acid and strong base are not equal. So, let's say we react 300 mL of a 1.0 mol solution of HCl with 100 mL of a 1.0 mol solution of sodium hydroxide, and our goal is to calculate the pH of the resulting solution. I should point out these are aqueous solutions of HCl and NaOH.

The first step is to calculate moles of strong acid and moles of strong base, and the equation for molarity is equal to moles over liters. For our strong acid HCl, the molarity is 1.0, and the volume is 300 mL, which is equal to 0.3 L. So, solving for x, we find that x is equal to 0.3 moles of HCl. For our strong base sodium hydroxide, the concentration is 1 molar, and the volume is 100 mL, which is equal to 0.1 L. So, solving for x, we find that x is equal to 0.1 moles of NaOH.

Going back to HCl, since HCl is a strong acid that ionizes 100%, if we have 0.3 moles of HCl, we have 0.3 moles of H⁺ ions, or we could say 0.3 moles of hydronium ions (H₃O⁺). For sodium hydroxide, since sodium hydroxide is a strong base that dissociates 100%, if we have 0.1 moles of sodium hydroxide, we have 0.1 moles of sodium ions and also 0.1 moles of hydroxide ions (OH⁻).

If we think about our net ionic equation, the 0.3 moles of hydronium ions will react with the 0.1 moles of hydroxide ions. So here's one way to write our net ionic equation: the hydronium ion plus the hydroxide ion forms 2 H₂O. To help us find the pH, we're going to use another ICF table. The initial moles of hydronium ions we calculate to be 0.3, and the initial moles of hydroxide ions we calculated to be 0.1.

Because we don't have equal amounts of moles and the mole ratio is 1:1, in this case, we're going to have a limiting reactant and an excess reactant. Since we only have 0.1 moles of hydroxide ions, all of the hydroxide ions will react. Since it's a 1:1 mole ratio, 0.1 moles of hydroxide ions will react with 0.1 moles of hydronium ions, so we can write minus 0.1 under hydronium as well.

Therefore, when the reaction goes to completion for the hydroxide ions, we'll have used all of them up. We started with 0.1, and we're using up 0.1, therefore we'll have 0 moles for the hydronium ion. We started with 0.3 moles minus 0.1 gives us 0.2 moles. So the hydroxide ions were our limiting reactant, and the hydronium ions were our excess reactant. The strong base neutralized some of the strong acid that was present, and we can calculate the pH of the resulting solution by the number of moles of excess hydronium ions.

Next, we calculate the concentration of hydronium ions in solution. Since the number of moles of hydronium ions is 0.2, we plug that into our equation for molarity. So, molarity is equal to moles over liters, and the total volume of our solution would be 300 mL plus 100 mL, which is 400 mL, or 0.4 L. So, let me go and write this in here. So 0.2 moles divided by 0.4 L is equal to a concentration of 0.5 mol.

Because the pH is equal to the negative log of the concentration of hydronium ions, if we plug in our concentration of hydronium ions of 0.5, the pH is equal to the negative log of 0.5, which is approximately 0.30. Having a low pH for our resulting solution at 25°C makes a lot of sense because we ended up with an excess of acid.