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Buffers

Pearson
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Buffers are solutions that can resist drastic changes in pH. Buffer solutions find great importance in chemistry, biology, and medicine. For example, our blood is a complex buffer, which keeps the pH at around 7.4 in our body. In these beakers, I have water, and you can see that when I add an acid and base, the pH will change greatly. So if I add a base to the solution, you can see that the pH increases quite significantly. Likewise, if I add an acid to this water solution, again the pH is lowered significantly. But buffers are resistant to pH changes, so what you think will happen when I add acid or base to these two beakers containing a buffer solution? So when I add a base to this buffer-containing solution, you can see on the pH meter that the pH does not change very much. The buffer is resistant to pH changes, and very similarly, if I add an acid to a buffer solution, you can see, again, that the pH does not change very much. The buffer maintains the pH even in the addition of an acid or a base. So why is that? Well, buffers contain a conjugate weak acid-based pair. What that means is we have both an acid and a base present. The acid present can neutralize the addition of any hydroxide ions, and the base present can neutralize the addition of any protons. So in these three beakers, I have different acid conjugate based combinations. Here I have hydrochloric acid and Cl-. Here I have HNO3 and NO3-, and here I have acetic acid and acetate ions. Which of these solutions could be a buffer? In this case, the acetic acid and acetate ion is the buffer. This is a weak acid-conjugate base pair. So if we look at the equation, we have our weak acid, HX, in equilibrium with H+ and our base X-. This is our weak acid, and here is our conjugate base, and these worke to neutralize the addition of any protons or hydroxide ions in our solution. If we look at the Ka expression, we simply have the products over the reactants, just as we've done. I can rearrange this and solve for H+. When I do that, I get the concentration of H+ is equal to the Ka times the concentration of HX, times the concentration of X-. So you can see that the concentration of the weak acid and the base is a ratio, and so with the addition of an acid or a base, I'm changing the concentration of X- and the HX in the solution, but you can see that there are ratio in this form. So the changes do not impact the concentration of H+, and hence, the pH very dramatically. So if I take the negative log of both sides of this equation, I get pH is equal to PKa minus log of HX over the concentration of X-. This is the Henderson-Hasselbalch equation, and you can see it directly relates the pH to the concentration of the acid and conjugate base in our buffer solution. So what would the pH of the solution of 0.6 molar acetic acid and 0.6 molar acetate ion be? So to solve this, we need the Ka, which we can find from the tables, and we can convert that to pKa. So my pH is equal to the negative log of my Ka minus the log of my concentration of the base and the acid, which in this case was 0.6 divided by 0.6. The Ka from the back of the book is 1.8 times 10 to the minus 5. So I have minus log of 1.8 times 10 to the minus 5, and the ratio here is 1. So we don't have to worry about this term. So in this case, the pH is just simply the pKa, which is equal to 4.74.
Buffers are solutions that can resist drastic changes in pH. Buffer solutions find great importance in chemistry, biology, and medicine. For example, our blood is a complex buffer, which keeps the pH at around 7.4 in our body. In these beakers, I have water, and you can see that when I add an acid and base, the pH will change greatly. So if I add a base to the solution, you can see that the pH increases quite significantly. Likewise, if I add an acid to this water solution, again the pH is lowered significantly. But buffers are resistant to pH changes, so what you think will happen when I add acid or base to these two beakers containing a buffer solution? So when I add a base to this buffer-containing solution, you can see on the pH meter that the pH does not change very much. The buffer is resistant to pH changes, and very similarly, if I add an acid to a buffer solution, you can see, again, that the pH does not change very much. The buffer maintains the pH even in the addition of an acid or a base. So why is that? Well, buffers contain a conjugate weak acid-based pair. What that means is we have both an acid and a base present. The acid present can neutralize the addition of any hydroxide ions, and the base present can neutralize the addition of any protons. So in these three beakers, I have different acid conjugate based combinations. Here I have hydrochloric acid and Cl-. Here I have HNO3 and NO3-, and here I have acetic acid and acetate ions. Which of these solutions could be a buffer? In this case, the acetic acid and acetate ion is the buffer. This is a weak acid-conjugate base pair. So if we look at the equation, we have our weak acid, HX, in equilibrium with H+ and our base X-. This is our weak acid, and here is our conjugate base, and these worke to neutralize the addition of any protons or hydroxide ions in our solution. If we look at the Ka expression, we simply have the products over the reactants, just as we've done. I can rearrange this and solve for H+. When I do that, I get the concentration of H+ is equal to the Ka times the concentration of HX, times the concentration of X-. So you can see that the concentration of the weak acid and the base is a ratio, and so with the addition of an acid or a base, I'm changing the concentration of X- and the HX in the solution, but you can see that there are ratio in this form. So the changes do not impact the concentration of H+, and hence, the pH very dramatically. So if I take the negative log of both sides of this equation, I get pH is equal to PKa minus log of HX over the concentration of X-. This is the Henderson-Hasselbalch equation, and you can see it directly relates the pH to the concentration of the acid and conjugate base in our buffer solution. So what would the pH of the solution of 0.6 molar acetic acid and 0.6 molar acetate ion be? So to solve this, we need the Ka, which we can find from the tables, and we can convert that to pKa. So my pH is equal to the negative log of my Ka minus the log of my concentration of the base and the acid, which in this case was 0.6 divided by 0.6. The Ka from the back of the book is 1.8 times 10 to the minus 5. So I have minus log of 1.8 times 10 to the minus 5, and the ratio here is 1. So we don't have to worry about this term. So in this case, the pH is just simply the pKa, which is equal to 4.74.