Strong acid solutions | Acids and bases | AP Chemistry | Khan Academy

A strong acid is an acid that ionizes 100% in solution. For example, hydrochloric acid (HCl) as a strong acid donates a proton to water (H2O) to form the hydronium ion (H3O+) and the conjugate base to HCl, which is the chloride ion (Cl−).

In reality, this reaction reaches an equilibrium; however, the equilibrium lies so far to the right and favors the product so much that we don't draw an equilibrium arrow. We simply draw an arrow going to the right, indicating the reaction essentially goes to completion. If the reaction essentially goes to completion, we can say that hydrochloric acid ionizes 100% and forms hydronium ions and chloride anions. So, essentially, there's no more HCl left; it's all turned into H3O+ and Cl−.

It's also acceptable to completely leave water out of the equation and to show hydrochloric acid (HCl) turning into H+ and Cl−. Once again, since HCl is a strong acid, there's only an arrow going to the right, indicating HCl ionizes 100%. And since there's only one water molecule difference between H+ and H3O+, H+ and H3O+ are used interchangeably.

Hydrochloric acid is an example of a monoprotic strong acid. Monoprotic means hydrochloric acid has one proton that it can donate in solution. Other examples of monoprotic strong acids include hydrobromic acid (HBr), hydroiodic acid (HI), nitric acid (HNO3), and perchloric acid (HClO4). Sulfuric acid is H2SO4, and it's a strong acid, but it's a diprotic acid, meaning it has two protons that it can donate. However, only the first ionization for sulfuric acid is strong.

Let's calculate the pH of a strong acid solution. In this case, we're going to look at a 0.4 mol solution of nitric acid. Nitric acid (HNO3) reacts with water to form hydronium (H3O+) and nitrate (NO3−), which is the conjugate base to HNO3. Because nitric acid is a strong acid, we assume the reaction goes to completion.

Therefore, if the initial concentration of nitric acid is 0.4 mol, looking at our mole ratio in the balanced equation, there's a one in front of nitric acid, and there's also a one in front of hydronium and a one in front of nitrate. Therefore, if the reaction goes to completion, the concentration of hydronium would also be 0.4 molar, and the same with the nitrate anion, that would also have a concentration of 0.4 mol.

Since our goal is to calculate the pH of this solution, we know that the equation for pH is pH = -log[H3O+]. Therefore, we just need to plug in the concentration of hydronium ions into our equation. This gives us pH = -log(0.4), which is equal to 1.40. So even though this is a pretty dilute solution of nitric acid, because nitric acid is a strong acid, the pH is pretty low.

Also note, since we have two significant figures for the concentration of hydronium ions, we need two decimal places for our final answer. Let's do another problem with a strong acid solution. Let's say we have 100 mL of an aqueous solution of hydroiodic acid, and the pH of the solution is equal to 1.50, and our goal is to find the mass of HI that's present in solution.

Hydroiodic acid reacts with water to form the hydronium ion and the iodide anion, and the mole ratio of HI to H3O+ is 1:1. So, if we can find the concentration of hydronium ion in solution, that should also be the initial concentration of hydroiodic acid. Once we find the initial concentration of hydroiodic acid, we can find the mass of HI that's present.

Since we're given the pH in the problem, we can plug that directly into our equation, which gives us 1.50 = -log[H3O+]. To solve for the concentration of hydronium ions, we can first move the negative sign to the left side, which gives us -1.50 = log[H3O+].

To get rid of the log, we can take 10 to both sides, so the concentration of hydronium ions is equal to 10^(-1.50), which is equal to 0.32. So, the concentration of hydronium ions is 0.32 molar, and because the mole ratio of hydronium ion to HI is 1:1, the initial concentration of HI is also 0.32 mol.

Now that we know the initial concentration of HI, we're ready to find the mass of HI present. Molarity is moles per liter, so let's go ahead and rewrite this as 0.32 moles per liter. The volume of the solution is 100 mL, which is equal to 0.10 L.

So, if we multiply moles per liter by the volume, which is 0.10 L, liters will cancel and give us moles. This is equal to 0.32 moles of HI. Since our goal is to find the mass of HI present, the final step is to multiply the moles of HI by the molar mass, which is 128 g per 1 mole of HI.

So, moles of HI would cancel out, and this gives us 41 grams as our final answer.