The percent ionization of a weak acid, HA, is defined as the ratio of the equilibrium H₃O⁺ concentration to the initial HA concentration, multiplied by 100%. In this video, we'll use this relationship to find the percent ionization of acetic acid in a 0.20 M solution. Created by Jay.
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- Am I getting the math wrong because, when I calculated the hydronium ion concentration (or X), I got 0.06x10^-3.(1 vote)
- It is funny when he excludes x because it is minor and easier to solve equation. I suggest we solve equations without deleting them. It is not that difficult.(1 vote)
- I mean yeah, we could solve the full equation since it’s only a quadratic. The benefit of using the small x approximation and the 5% rule is that it saves us time calculating the real answer compared to using the quadratic formula. Similar to how estimating numbers in multiplication gives us an idea of the magnitude of the real answer.
Additionally, the reason the 5% rule is allowed is because the difference between the real answer and the approximate answer is insignificant. Especially when we round for significant figures, often the answers will be identical. So you can put in this extra work using the quadratic formula and get the same answer as someone using this approximation.
Ultimately, you get the same answer. It’s just a matter of saving time.
Hope that helps.(1 vote)
- [Instructor] Let's say we have a 0.20 Molar aqueous solution of acidic acid. And our goal is to calculate the pH and the percent ionization. The Ka value for acidic acid is equal to 1.8 times 10 to the negative fifth at 25 degrees Celsius. First, we need to write out the balanced equation showing the ionization of acidic acid. So acidic acid reacts with water to form the hydronium ion, H3O+, and acetate, which is the conjugate base to acidic acid. Because acidic acid is a weak acid, it only partially ionizes. Therefore, we need to set up an ICE table so we can figure out the equilibrium concentration of hydronium ion, which will allow us to calculate the pH and the percent ionization. In an ICE table, the I stands for initial concentration, C is for change in concentration, and E is equilibrium concentration. The initial concentration of acidic acid is 0.20 Molar. So we can put that in our ICE table under acidic acid. And if we assume that the reaction hasn't happened yet, the initial concentrations of hydronium ion and acetate anion would both be zero. Some of the acidic acid will ionize, but since we don't know how much, we're gonna call that x. So we write -x under acidic acid for the change part of our ICE table. And when acidic acid reacts with water, we form hydronium and acetate. So we're going to gain in the amount of our products. To figure out how much we look at mole ratios from the balanced equation. There's a one to one mole ratio of acidic acid to hydronium ion. Therefore, if we write -x for acidic acid, we're gonna write +x under hydronium. And for the acetate anion, there's also a one as a coefficient in the balanced equation. Therefore, we can write +x under acetate as well. So the equilibrium concentration of acidic acid would be 0.20 minus x. Let's go ahead and write that in here, 0.20 minus x. The equilibrium concentration of hydronium would be zero plus x, which is just x. And for acetate, it would also be zero plus x, so we can just write x here. Next, we brought out the equilibrium constant expression, which we can get from the balanced equation. So the Ka is equal to the concentration of the hydronium ion. And since there's a coefficient of one, that's the concentration of hydronium ion raised to the first power, times the concentration of the acetate anion also raised to the first power, divided by the concentration of acidic acid raised to the first power. And water is left out of our equilibrium constant expression. Because the concentrations in our equilibrium constant expression or equilibrium concentrations, we can plug in what we have from our ICE table. So we can plug in x for the equilibrium concentration of hydronium ion, x is also the equilibrium concentration of the acetate anion, and 0.20 minus x is the equilibrium concentration of acidic acid. We also need to plug in the Ka value for acidic acid at 25 degrees Celsius. Here we have our equilibrium concentrations plugged in and also the Ka value. Our goal is to solve for x, which would give us the equilibrium concentration of hydronium ions. However, if we solve for x here, we would need to use a quadratic equation. So to make the math a little bit easier, we're gonna use an approximation. We're gonna say that 0.20 minus x is approximately equal to 0.20. The reason why we can make this approximation is because acidic acid is a weak acid, which we know from its Ka value. Ka is less than one. And that means it's only going to partially ionize. It's going to ionize to a very small extent, which means that x must be a very small number. And if x is a really small number compared to 0.20, 0.20 minus x is approximately just equal to 0.20. So we can go ahead and rewrite this. So we would have 1.8 times 10 to the negative fifth is equal to x squared over, and instead of 0.20 minus x, we're just gonna write 0.20. Solving for x, we would find that x is equal to 1.9, times 10 to the negative third. And remember, this is equal to the equilibrium concentration of hydronium ions. So let's write in here, the equilibrium concentration of hydronium ions. So this is 1.9 times 10 to the negative third Molar. If we would have used the quadratic equation to solve for x, we would have also gotten 1.9 times 10 to the negative third to two significant figures. Therefore, using the approximation got us the same answer and saved us some time. Also, now that we have a value for x, we can go back to our approximation and see that x is very small compared to 0.20. So 0.20 minus x is approximately equal to 0.20. Also, this concentration of hydronium ion is only from the ionization of acidic acid. And it's true that there's some contribution of hydronium ion from the autoionization of water. However, that concentration is much smaller than this. So for this problem, we can ignore the contribution of hydronium ions from the autoionization of water. Next, we can find the pH of our solution at 25 degrees Celsius. So pH is equal to the negative log of the concentration of hydronium ions. So that's the negative log of 1.9 times 10 to the negative third, which is equal to 2.72. We also need to calculate the percent ionization. So the equation 4% ionization is equal to the equilibrium concentration of hydronium ions, divided by the initial concentration of the acid, times 100%. The equilibrium concentration of hydronium ions is equal to 1.9 times 10 to negative third Molar. So we plug that in. And the initial concentration of our weak acid, which was acidic acid is 0.20 Molar. So the Molars cancel, and we get a percent ionization of 0.95%. A low value for the percent ionization makes sense because acidic acid is a weak acid. We can also use the percent ionization to justify the approximation that we made earlier using what's called the 5% rule. So let me write that down here, the 5% rule. If the percent ionization is less than 5% as it was in our case, it was less than 1% actually, then the approximation is valid. If the percent ionization is greater than 5%, then the approximation is not valid and you have to use the quadratic equation.