Auto-Ionization - Online Tutor, Practice Problems & Exam Prep

Water undergoes self-ionization, producing hydronium (H₃O⁺) and hydroxide (OH^-) ions. This process follows the Brønsted–Lowry theory, where one water molecule acts as an acid, donating H⁺, while the other acts as a base. The ion product constant for water, K_w, is defined as K_w = [H⁺][OH^-], with a standard value of 1.0 × 10^-14 at 25°C. This relationship allows for calculations involving acid-base reactions and equilibrium expressions, emphasizing the temperature dependence of K_w.

Auto-Ionization

Now water can react with another water molecule in a process called self-ionization or autoionization. In the process, we create H₃O⁺ (hydronium ion) and OH^- (hydroxide ion). Here, one water molecule acts as an acid while the other acts as a base. Following the Bronsted-Lowry theory, acids donate H⁺. So, this acid donates an H⁺. In the process, the water becomes OH^- because it's lost an H⁺. Charged ions exist as aqueous species within water; plus, the other water accepts an H⁺ and becomes H₃O⁺ aqueous. Now, this is the actual reaction: the 2 water molecules reacting with one another to create those two products.

We can simplify this by simply saying we have water as a liquid, and then we have the creation of H⁺ ion and OH^- ion. In reality, the water molecule isn't splitting into these ions because that's not how water works. This is just a simplified explanation of what's occurring. From these two equations, we should realize that H⁺ and H₃O⁺ are equal to one another; they're the same exact thing. H⁺ is just a simplified representation of the hydronium ion of H₃O⁺.

We're going to say that with these equations, we can talk about an equilibrium expression. Here, we're going to say the equilibrium constant for water is called the ion product constant or simply Kw. Just like all other equilibrium constants, it equals products over reactants. So whether we're looking at the first equation or the second equation, it really doesn't matter. In both, it's products over reactants; remember, we ignore solids and liquids within our equilibrium expression. Here we have liquid so they would be ignored. If we look at the simplified reaction, we ignore this liquid so reactants wouldn't be taken into account at all. So Kw just simply breaks down to equal products. It'd be equal to H⁺ concentration times OH^- concentration. This is how we get the development of this equation here. Because of this equation, we can draw comparisons and calculations dealing with H⁺ concentration and OH^- concentration. If we know one, we know the other because Kw, our ion product constant, will serve as a constant within our calculations.

Now, with most calculations, we're going to be dealing with it under normal or laboratory conditions. That means the temperature will be 25 degrees Celsius. At 25 degrees Celsius, Kw is 1.0 times 10^-14, a value you'll need to commit to memory. Now, like all other equilibrium constants, it's susceptible to changes when we manipulate the temperature. So Kw, just like all other equilibrium constants, is heavily temperature-dependent. If you change the temperature, you'll change what the new Kw value will be. If they do change the temperature, they would have to give you a new Kw value because there's no way of you really knowing. The only temperature that you're supposed to remember is at 25 degrees Celsius. This is the established value for Kw. And remember, Kw is what connects my hydronium ion concentration to my hydroxide ion concentration. Taking these into account, we'll attempt to do the 2 example questions left on the bottom of the page. So click on to the next video and see how I approach example 1 in terms of relating Kw to H⁺OH^-.

Auto-Ionization

Here it states at 0 degrees Celsius the kw for a neutral solution is recorded as 1.2×10-15. Based on what you've reviewed, what can be said in terms of kw and the solution?

Alright, so we know that kw is temperature-dependent as mentioned above. We know that at 25 degrees Celsius, kw equals 1.0×10-14. And here they just told us at 0 degrees Celsius, kw equals 1.2×10-15. Alright. So, if we just think about this neutral solution, think of it as we have H2O liquid plus H2O liquid.

These molecules would react with one another under auto-ionization or self-ionization to produce H3O+ aqueous and OH- aqueous. Here, the temperature is increasing from 0 to 25 degrees Celsius, so our temperature increased. According to Le Châtelier's principle, it states that if we increase our temperature, we shift away from heat. And here our kw goes from 10-15 to 10-14. So, our kw has increased.

kw, just like all other equilibrium constants k, is equal to products over reactants. If your kw is increasing, products are favored, and the only way products can be favored is if we're moving towards them. So we're moving this way towards my products, causing them to increase. And again, according to Le Châtelier's principle, if we increase the temperature, we're going to shift away from heat. Since we're shifting to the right, that must mean that heat is here on the left. And if heat is a reactant, that means heat is being absorbed, so this would be an endothermic process.

Again, this is taking what we've learned in terms of acids and bases in reference to kw, but applying concepts we've seen before with Le Châtelier's principle. Again, increasing temperature, decreasing temperature means we're going to apply the rules we've learned about Le Châtelier's principle to find out if our reaction is truly exothermic or endothermic. Recall that what I've said in earlier videos: when we're talking about thermal neutral, that means that delta H will be equal to 0. There'll be no real change in temperature. Therefore, there will be no shifting left or right because your kw or all equilibrium constants would not change under thermal neutral conditions. And we know that we have enough information to answer the question. We determined it was endothermic, not exothermic. So based on what we've learned thus far, look to see if you can attempt example 2. If you get stuck, don't worry. Just click on to the next video and see how I approach that same question.

Auto-Ionization

Here it states: determine the concentration of hydronium ions for pure water at 50 degrees Celsius. Here again, temperature is changing so our \( K_w \) will be different from what we're used to seeing. \( K_w \) now is \( 5.3 \times 10^{-14} \) because again \( K_w \) is temperature dependent. Changing the temperature changes \( K_w \) from the common number of \( 1.0 \times 10^{-14} \) to a new value. Now, here they're telling us that we have pure water. Pure water means that it is neutral. Neutral water means that \( OH^- \) concentration is equal to \( H^+ \) concentration. And since we don't know what either one is, you say they're both equal to \( x \). They're connected by the equation of \( K_w = [H^+][OH^-] \). So both of them are equal to \( x \), so \( x \times x \) gives us \( x^2 \). They're both equal to this new value for \( K_w \) since the temperature is not 25 degrees Celsius. So that’s \( 5.3 \times 10^{-14} \). We're looking for hydronium ion concentration, which is \( H^+ \). So, we just need to isolate \( x \). So take the square root of both sides here. When we do that, that's going to give me my concentration for \( x \). When I do that, I get \( x = 2.30 \times 10^{-7} \) molar. Since it's neutral, this would be the concentration of \( H^+ \) as well as \( OH^- \) in this question. But remember, at 25 degrees Celsius, \( K_w \) is \( 1.0 \times 10^{-14} \). Once you change the temperature, it becomes a brand new value. You'd be given that new value. The only one that you need to memorize is the one at 25 degrees Celsius. So just remember the connection between hydronium ion and hydroxide ion in reference to the ion product constant of \( K_w \).

Here’s what students ask on this topic:

What is the auto-ionization of water?

The auto-ionization of water is a process where two water molecules react to form hydronium (H₃O⁺) and hydroxide (OH^-) ions. This reaction can be represented as:

$H_2\left(l\right)+H_2\left(l\right)\to H_{3}O+\left(\mathrm{aq}\right)+{\mathrm{OH}}_{-}$

In this process, one water molecule acts as an acid, donating an H⁺ ion, while the other acts as a base, accepting the H⁺ ion. This equilibrium is characterized by the ion product constant for water, K_w, which is 1.0 × 10^-14 at 25°C.

What is the ion product constant for water (Kw) and how is it used?

The ion product constant for water, denoted as K_w, is the equilibrium constant for the auto-ionization of water. It is defined as:

$K_w=\left[H_{+}\right]\left[{\mathrm{OH}}_{-}\right]$

At 25°C, K_w is 1.0 × 10^-14. This constant allows us to calculate the concentrations of H⁺ and OH^- ions in water. If we know the concentration of one ion, we can determine the other using the relationship:

$K_w=\left[H_{+}\right]\left[{\mathrm{OH}}_{-}\right]$

This is crucial for understanding acid-base reactions and pH calculations.

How does temperature affect the ion product constant (Kw) of water?

The ion product constant for water, K_w, is temperature-dependent. At 25°C, K_w is 1.0 × 10^-14. However, as temperature increases, K_w also increases because the auto-ionization of water is an endothermic process. Conversely, as temperature decreases, K_w decreases. This means that the concentrations of H⁺ and OH^- ions in water will vary with temperature. For accurate calculations involving K_w at temperatures other than 25°C, the specific K_w value at that temperature must be provided.

What is the relationship between H⁺ and H₃O⁺ in the context of auto-ionization of water?

In the context of the auto-ionization of water, H⁺ and H₃O⁺ are often used interchangeably. H⁺ represents a proton, which does not exist freely in water but associates with a water molecule to form the hydronium ion, H₃O⁺. Therefore, H⁺ is a simplified notation for H₃O⁺. The reaction can be represented as:

$H_2\left(l\right)+H_2\left(l\right)\to H_{3}O+\left(\mathrm{aq}\right)+{\mathrm{OH}}_{-}$

Thus, H⁺ and H₃O⁺ are essentially the same in aqueous solutions.

Why is the value of Kw important in acid-base chemistry?

The value of K_w is crucial in acid-base chemistry because it defines the relationship between the concentrations of H⁺ and OH^- ions in water. This relationship is given by:

$K_w=\left[H_{+}\right]\left[{\mathrm{OH}}_{-}\right]$

At 25°C, K_w is 1.0 × 10^-14. This constant allows us to calculate the pH and pOH of solutions, which are measures of acidity and basicity, respectively. Understanding K_w helps in predicting the behavior of acids and bases in water, making it fundamental for solving equilibrium problems and understanding the nature of aqueous solutions.