pH Revisited - Online Tutor, Practice Problems & Exam Prep

pH is calculated as the negative logarithm of the hydronium ion concentration (H₃O⁺). In pure water at 50°C, the ion product constant (K_w) is 7.94 × 10⁻¹⁴, leading to a pH of approximately 6.55. When free ions are present, ionic strength and activity coefficients must be considered, affecting the pH calculation. For example, in a saturated barium hydroxide solution with lithium nitrite, the pH is influenced by the ionic strength, resulting in a pH of about 13.41 after accounting for activity coefficients.

Glass Electrodes

Remember that pH is just the negative log of H⁺ or H₃O⁺. For example one, it asks us if at 50 degrees Celsius, the ionization of pure water Kw is 7.94 times 10 to the negative 14. What is the pH of a neutral solution? Well, for a question like this, we're dealing with pure water, with no free floating ions within it. Here, we're dealing with a neutral solution.

And in a neutral solution, your H⁺ concentration is equal to your OH^- concentration. Since we don't know either one of them, we're going to say they're both equal to x. Remember, both of these ions are related to each other by the fact that H⁺ or H₃O⁺ times OH^- equals Kw. And here, since both are x and they're multiplying, that's going to be x squared. That equals 7.94 times 10 to the negative 14.

Remember, your Kw, like all other equilibrium constants, is temperature dependent. Once the temperature is no longer 25 degrees Celsius, our Kw will not be 1.0 times 10 to the negative 14. It becomes a brand new number. Here, we're going to take the square root of both sides to isolate x. That's going to give me 2.8178 times 10 to the negative 7 molar.

That equals H⁺ as well as OH^-. But since we're looking at pH, we're going to focus only on the H⁺. pH equals negative log of H⁺ which equals negative log of 2.8178 times 10 to the negative 7. That gives me 6.55 as my concentration after rounding. That gives us option B. Now, that's when we're dealing with pure water.

What would happen when we're dealing with figuring out the pH still in pure water, but now it has free floating ions in it? We have to take into account what does this do to both our ionic strength as well as our activity coefficient. Think of those things as you read example 2. You could attempt it on your own but if you get stuck, don't worry. Just come back and see how I approach that same example 2 question.

Glass Electrodes

Alright. Here we're dealing with the same pure water. We're going to say because it's the same pure water, that means that H⁺ still equals OH^-. And they both still equal x. But now we're dealing with free-floating ions as well.

Now Na⁺Br^- have nothing to do with H⁺ and OH^-. And because they don't have a common ion effect involved, that means they're going to have some form of ionic strength and some influence on these 2 ions. So again, if they're giving you ionic compounds and those ions don't match up with the ions that you're investigating, that means ionic strength is in play. And if ionic strength is in play, that means that our activity coefficients are also in play. Here, our equation becomes a bit different.

It becomes H+8⁢γOH-=kw. We're still dealing with this 50 degrees Celsius from earlier, so we're still dealing with the same k_w. Now we know that these 2 are x, but we need to know what the activity coefficients are for them in order to figure out our correct H⁺ ion concentration. Now to do that, we're gonna figure out what the ionic strength of sodium bromide is. Once we get that ionic strength, we're gonna look it up on the activity coefficients table and see what the activity coefficients are for H⁺ and OH^-.

Plug those in to find our final answers. We're going to need room, guys. Let me take myself out of the image. We're going to say here ionic strength equals. This breaks up into Na⁺1 and Br^-1.

This is half the concentration of each ion times charge squared plus the concentration of the other ion times charge squared. The ionic strength is 0.05. Look on the activity coefficients table. Look for H⁺ ion and OH^-. What are their values when the ionic strength is this number?

If you look carefully, you'll see that for H⁺ it's 0.933. And for OH^- it's 0.926. Remember k_w is the same k_w that we saw previously which is 7.94x10^-14 when the temperature is 50 degrees Celsius. We're gonna plug that in. Now we're going to multiply these two numbers together.

So when we multiply both of them together, it gives me 0.863958. And then we have x variables multiplying, so that's x². Now, we're going to divide both sides by 0.863958. That's going to give me x². x² equals 9.19x10^-14.

Take the square root of both sides here. So that gives me x, x equals 3.03 times 10 to the negative 7 molar. That's gonna give me the concentration of both H⁺ and OH^-. But we're only concerned with the H⁺ because we're looking for pH. So up above, we said pH equaled negative log of H⁺.

But remember, we're dealing with ionic strength and because of that, we're going to have to deal with the activity coefficient. Input the activity coefficient of H⁺ again. Now we're going to do negative log. The concentration is 3.03 times 10 to the negative 7 times the activity coefficient of H⁺ which is 0.933. What's happening here is you're going to multiply these two numbers together and then you're going to take, the negative log of that because remember, both of these things together represent the activity of the H⁺ ion.

When I multiply those two numbers together, I get 2.828x10^-7 which equals 6.54. K. So our activity coefficient had a small impact on the pH where initially I got 6.55 for my pH, I now have gotten 6.54. Realize here that if we increased the concentration and or if we increased the charges of the ions, the impact on the pH would have been much greater than just a difference of 0.01. This just goes to show you that if we have non-common ions floating around, we have to take into account ionic strength and therefore the activity coefficients of the ions involved.

Now that we've seen this, take a look at example 3 left on the bottom. The same basic premise is being involved here. We're dealing with the Ksp of barium hydroxide, but we have involved lithium nitrate, non-common ions. Again, that means that your ionic strength is in play, which means activity coefficients are in play. Remember what we just did and see if you can solve this question.

Glass Electrodes

Here it says, find the pH of a saturated solution of barium hydroxide when dissolved in 0.05 molar of lithium nitrite. Here we're dealing with molar lithium nitrite. Here the Ksp of barium hydroxide is 5.0×10-3. Well, we know that we're dealing with barium hydroxide. This is my ionic solid.

We're talking about how that ionic solid breaks up into its ions. We know that we don't deal with solids. Here, this would be x. And here, because there's a 2 here, this would be 2x. Ksp equals just products because we're ignoring the solid that's the reactant.

We have Ba2+. Now realize here that because we're dealing with non-common ions, that means ionic strength is in play and therefore the activity coefficient is in play. So it's going to be activity coefficient of barium 2 ion times hydroxide ion. Because of the 2 here, it's squared times the activity coefficient of hydroxide also squared. We're going to need to determine what our concentration of hydroxide ion is to figure out what our pOH is.

And once we do that, we can find our pH. But first, I need to take into account these non-common ions. From it, I'll be able to determine my ionic strength. This is made up of lithium ion and nitrate ion. It's a one-to-one relationship, so the concentrations will not change.

0.05 molar for the lithium ion times its charge squared plus the concentration of the nitrate ion times its charge squared. That gives me 0.05 for the ionic strength.

Now that we know what the ionic strength is, we look up on our activity coefficients chart for the activity coefficients for barium ion and hydroxide ion. So when you look those up, so we have barium ion here which is x. When you look up at its activity coefficient, you're going to see that it is equal to 0.465 times hydroxide which is 2x.

Don't forget that it's squared Times the activity coefficient. When you look it up, you get 0.81 squared. Ksp is 5.0×10-3 equals all right. We have a lot of numbers that are multiplying with one another. We're going to have to take all of those into account.

We have 0.465 times 0.81 squared. 2 squared is 4. When you multiply all those together, that gives me 1.22035. Don't forget, we have our x variables. We have x here and then x here is getting squared so that's x squared.

X times x squared gives me x cubed. Divide 1.22035. So x cubed equals 0.004097. Take the cube root of both sides here. So when we do that, that's going to give me x equals 0.160016 molar.

But remember, we're looking for the concentration of hydroxide ion, OH-. And OH- does not equal x. It equals 2x. So OH- concentration equals 2x. It's 2 times this number we just found.

That's 0.320032 molar. Next, we're going to figure out what the activity of hydroxide is. The activity here would equal the concentration of the hydroxide ion we just found times its activity coefficient. The concentration we just found is this. And we bring back that activity coefficient we found earlier of 0.81.

That's going to give me 0.259225. Now that I have the activity of hydroxide, I can take the negative log of that to find pOH. So that's negative log of that number. So that's gonna give me a pOH of 0.586 pH equals 14 minus pOH. My pH at the end is approximately 13.41.

Again, remember, when we're dealing with an ionic compound that's dissolving in solution and then all of a sudden they introduce non-common ion effect ions, that means that those are going to help increase the solubility of my ionic compound by ionic strength. Since ionic strength is in play, that means activity coefficients are in play. We're going to have to incorporate them within our calculations to find the final concentrations of the specified ions. Keep in mind some of the techniques we've used here, which is just a continuation of concepts we've learned in terms of calculating pH.