Worked examples: Calculating [H₃O⁺] and pH | Acids and bases | AP Chemistry | Khan Academy

• [Instructor] Here are some equations that are often used in pH calculations. For example, let's say a solution is formed at 25 degrees Celsius and the solution has a pOH of 4.75, and our goal is to calculate the concentration of hydronium ions in solution, H3O+.

One way to start this problem is to use this equation, pH plus pOH is equal to 14.00. And we have the pOH equal to 4.75, so we can plug that into our equation. That gives us pH plus 4.75 is equal to 14.00. And solving for the pH, we get that the pH is equal to 9.25.

So we have the pH, and our goal is to solve for the concentration of hydronium ions, and pH is equal to the negative log of the concentration of hydronium ions. So we can plug our pH right into this equation. So that would give us the pH, which is 9.25, is equal to the negative log of the concentration of hydronium ions.

Next, we need to solve for the concentration of hydronium ions. So we could move the negative sign over to the left side, which gives us negative 9.25 is equal to the log of the concentration of hydronium ions. And to get rid of that log, we can take 10 to both sides.

So that's gonna give us the concentration of hydronium ions, H3O+, is equal to 10 to the negative 9.25. And 10 to the negative 9.25 is equal to 5.6 times 10 to the negative 10. So the concentration of hydronium ions in our solution at 25 degrees Celsius is equal to 5.6 times 10 to the negative 10th molar.

Also notice that because we had two decimal places for our pH, we have the concentration of hydronium ions to two significant figures. There's another way to do the same problem. Since we have the pOH, we could use the pOH equation to find the concentration of hydroxide ions in solution.

So we would just need to plug in the pOH into this equation which gives us 4.75, which is the pOH, is equal to the negative log of the concentration of hydroxide ions. Next, we can move the negative sign over to the left side. So negative 4.75 is equal to the log of the concentration of hydroxide ions.

And to get rid of the log, we just take 10 to both sides. 10 to the negative 4.75 is equal to 1.8 times 10 to the negative fifth. So the concentration of hydroxide ions is equal to 1.8 times 10 to the negative fifth molar.

So our goal is to find the concentration of hydronium ions in solution, and we have the concentration of hydroxide ions in solution, so we can use the Kw equation because the concentration of hydronium ions times the concentration of hydroxide ions is equal to Kw, which is equal to 1.0 times 10 to the negative 14th at 25 degrees Celsius.

So we can plug in the concentration of hydroxide ions into our equation. That gives us 1.8 times 10 to the negative fifth times the concentration of hydronium ions, which we'll just write as x in our equation is equal to Kw, which is equal to 1.0 times 10 to the negative 14.

Solving for x, we find that x is equal to 5.6 times 10 to the negative 10th. So the concentration of hydronium ions is equal to 5.6 times 10 to the negative 10th molar. So even though we used two different equations from the first time we did this problem, we ended up with the same answer that we did the first time, 5.6 times 10 to the negative 10th molar.

So it doesn't really matter which approach you take. Finally, let's look at an example where the temperature is not 25 degrees Celsius. Let's say we have a sample of pure water at 50 degrees Celsius, and our goal is to calculate the pH.

Pure water is a neutral substance, which means the concentration of hydronium ions, H3O+, is equal to the concentration of hydroxide ions, OH-. Right now, we don't know what those concentrations are, but we know that the concentration of hydronium ions times the concentration of hydroxide ions is equal to Kw.

However, we have to be careful because Kw is only equal to 1.0 times 10 to the negative 14th at 25 degrees Celsius. And at 50 degrees Celsius, Kw is equal to 5.5 times 10 to the negative 14th. So let's go ahead and write Kw is equal to 5.5 times 10 to the negative 14th.

And if we make the concentration of hydronium ions x, then the concentration of hydroxide ions would also have to be x since the two are equal. So we would have x times x is equal to 5.5 times 10 to the negative 14th.

So that gives us x squared is equal to 5.5 times 10 to the negative 14th. And if we take the square root of both sides, we find that x is equal to 2.3 times 10 to the negative seventh.

And since x is equal to the concentration of hydronium ions in solution, the concentration of hydronium ions is 2.3 times 10 to the negative seventh molar. Now that we know the concentration of hydronium ions in solution, we can use our pH equation to find the pH of water at 50 degrees Celsius.

So we plug our concentration of hydronium ions into our equation, and we take the negative log of that, and we get that the pH is equal to 6.64. Notice, with two significant figures for the concentration, we get two decimal places for our answer.

If we had used Kw is equal to 1.0 times 10 to the negative 14th, we would've gotten a pH of 7.00, but that's only true at 25 degrees Celsius. Since Kw is temperature dependent, if the temperature is something other than 25 degrees Celsius, the pH of water will not be seven.

So in this case, we calculated it to be 6.64. The pH of pure water is 6.64 at 50 degrees Celsius. However, water is still a neutral substance. The concentration of hydronium ions is equal to the concentration of hydroxide ions.