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AP.Chem:
SAP‑9 (EU)
,
SAP‑9.C (LO)
,
SAP‑9.C.3 (EK)
,
SAP‑9.C.4 (EK)
,
SAP‑9.C.5 (EK)

Video transcript

- [Instructor] An example of a weak base is ammonia, NH3, and ammonia will react with water. And in this reaction, water functions as a Bronsted-Lowry acid and donates a proton to ammonia, which functions as a Bronsted-Lowry base and accepts a proton. A proton is H plus. So if you add on H plus onto NH3, you form NH4 plus, which is the ammonium ion. And if you take away NH plus from H2O, you form the hydroxide anion, which is OH minus. This reaction comes to an equilibrium, so we have an equilibrium arrow in here, and we can also write an equilibrium constant expression. So first, we look at our products. We have NH4 plus, and there's a coefficient of one in front of it, so it'd be the concentration of NH4 plus raised to the first power times the concentration of the other product, hydroxide, which also has a coefficient of one, so it's the concentration of hydroxide raised to the first power, divided by the concentration of ammonia, and the coefficient is also one, so that's also raised to the first power. And water is left out of our equilibrium constant expression. The equilibrium constant K has a subscript of b, which stands for base. And this Kb is called the base ionization constant, or the base dissociation constant. Ammonia is a weak base, which means it only partially ionizes in water. And if it only partially ionizes, at equilibrium, there's going to be a small amount of products and a large amount of reactants. Therefore, looking at the equilibrium constant expression, the numerator is gonna be very small compared to the denominator. So with a small numerator and a large denominator, that means that the Kb value for a weak base will be less than one. Let's look at a few examples of weak bases. Here's the dot structure for ammonia, and notice how ammonia has a nitrogen with a lone pair of electrons that's capable of accepting a proton. A proton is H plus. So if we add an H plus onto NH3, we form NH4 plus, which is the ammonium ion, and it's the conjugate acid to ammonia. So think about this lone pair of electrons that I circled here picking up a proton to form this bond. If we replace the hydrogen with what's called a methyl group, CH3, we form a new weak base called methylamine. Like ammonia, methylamine has a nitrogen with a lone pair of electrons that's capable of accepting a proton. So if methylamine picks up a proton, it forms the methyl ammonium ion. Both ammonia and methylamine are examples of neutral substances. However, a weak base doesn't have to be neutral. If we look at the hypochlorite anion, ClO minus, if the hypochlorite anion picks up a proton, let's say this lone pair of electrons in the oxygen picks up a proton, it would form hypochlorous acid, and hypochlorous is the conjugate acid to hypochlorite. So these are three examples of weak bases, which we can tell by looking at their Kb values at 25 degrees Celsius. Notice they all have Kb values less than one. However, the higher the Kb value, the higher the equilibrium concentration of hydroxide ion, and therefore the stronger the base. So out of these three week bases, the one with the higher Kb value is the most basic. So looking at these three values, again, the Kb value for methylamine is the highest of the three, so it's the strongest base of the three, although all three are considered to be weak. Let's do a calculation with a weak base. Let's say we have a .14 molar solution, aqueous solution of ammonia at 25 degrees Celsius, and our goal is to calculate the concentration of hydroxide ions in solution, and also the pH of the solution at 25 degrees Celsius. So first, we're gonna calculate the equilibrium concentration of hydroxide ions, and to help us do that, we're gonna use an ICE table, where I stands for the initial concentration, C is the change in concentration, and E is the equilibrium concentration. The initial concentration of ammonia is .14 molar. So we can put that in on our ICE table. And if we assume that ammonia hasn't ionized at all yet, the initial concentrations of ammonium and hydroxide ions would both be zero. Next, we think about some of the ammonia ionizing, and we don't know how much, so we write minus x under ammonia for the change. When NH3 gains a proton, it turns into NH4 plus, and the mole ratio of ammonia to ammonium is one to one. Therefore, if we write minus x for ammonia, we must write plus x for ammonium. And the hydroxide ion also has a one as a coefficient in a balanced equation. Therefore, it would also be plus x in our ICE table. So the equilibrium concentration of ammonia would be .14 four minus x. So let's write that in there, .14 minus x. For ammonium, it would be zero plus x, which is just x, and for hydroxide, it would also be zero plus x, which is just x. The next step is to write out the equilibrium constant expression for the ionization of ammonia, which we've already talked about. And since the concentrations in our equilibrium constant expression are equilibrium concentrations, we can go ahead and plug in the equilibrium concentrations of the ammonium ion, the hydroxide ion, and ammonia into our equilibrium constant expression. We also need to plug in for Kb for ammonia at 25 degrees Celsius. Here's our equilibrium constant expression with everything plugged in. And our goal is to solve for x, because x represented the equilibrium concentration of hydroxide ions. However, if we solve for x now, we would get a quadratic equation. So to simplify the math, we're gonna use an approximation. We're gonna say that .14 minus x is approximately equal to .14. The reason why we can make this approximation is because the Kb value for ammonia is quite small, which means that ammonia doesn't ionize very much at all, which tells us the equilibrium concentration of hydroxide ions will be a very small number. And if x is a really small number compared to .14, .14 minus x is just approximately equal to .14. So instead of writing .14 minus x, we can just write in .14. So now we solve for x and we find that x is equal to 1.6 times 10 to the negative third molar, which is the equilibrium concentration of hydroxide ions. Our goal is to calculate the pH of our solution. And first, we could find the pOH, because pOH is equal to the negative log of the concentration of hydroxide ions. So the negative log of 1.6 times 10 to the negative third is equal to 2.80. And at 25 degrees Celsius, the pH plus the pOH is equal to 14.00, so we could plug in the pOH into our equation and solve for the pH, and we would find the pH of our solution is equal to 11.20. And this makes sense, because we have a basic solution, so the pH should be greater than seven. Finally, to make sure that we can use our approximation, let's find the percent ionization for our weak base. To calculate the percent ionization of ammonia, we take x, which you could think about as the concentration of ammonia that ionized, or you could think about it as the equilibrium concentration of hydroxide ions. And we divide x by the initial concentration of ammonia and multiply it by 100 to get a percent ionization. So for this problem, x is equal to 1.6 times 10 to the negative third molar. So we go ahead and plug that in here. So 1.6 times 10 to the negative third molar, and the initial concentration of ammonia was equal to .14 molar. So molar cancels out and we end up with a percent ionization of 1.1%. Once you calculate the percent ionization, you need to think about the 5% rule. And the 5% rule says, if your percent ionization is less than 5%, it's okay to use the approximation. So in our case, our percent ionization was 1.1%, which is less than 5%. So it's okay to say that .14 minus x is approximately equal to .14, because with such a low percent ionization, x is a really small value compared to .14. However, if you get a percent ionization greater than 5%, you can't use the approximation and you need to use the quadratic equation to calculate x. Finally, when we calculated our equilibrium concentration of hydroxide ions, we only considered the ionization of ammonia. It's true that the autoionization of water would produce a small amount of hydroxide ions in solution. However, that concentration is extremely small compared to the concentration of hydroxide ions from the ionization of ammonia. Therefore, we ignore the contribution of hydroxide ions from the auto ionization of water in our calculations.
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