Titrations of Acids and Bases, Part 1

Need help preparing for the General Chemistry section of the MCAT? MedSchoolCoach expert, Ken Tao, will teach Part 1 about Titrations of Acids and Bases. Watch this video to get all the MCAT study tips you need to do well on this section of the exam!
If you’ve ever enrolled in an introductory chemistry lab, you likely performed an experiment where you pipetted a liquid into a solution until the solution changed color. You were performing a titration reaction, which is an essential experimental technique on the MCAT. Titrations are a great tool for determining the concentration or molar amount of an unknown acid or base in a solution. They are performed by measuring the pH of an unknown solution while adding a strong acid or base until the solution reaches its equivalence point, at which time the pH will change dramatically. By graphing the changes in pH against the amount of strong acid/base added, we can triangulate several key points that give us information about the solution.
Understanding the Graph of a Titration
On the MCAT, the unknown solution we want to understand is often called the analyte, and the strong acid or base we use to titrate it is often called the titrant. If we found a beaker in a laboratory filled with an unknown analyte, the first step in performing a titration would be figure out if we want to use an acidic or basic titrant. In order to do this, we would put a pH probe in the solution and note whether the pH is above seven or below seven. If it's below seven, the unknown solution is acidic, and so the titration will involve adding a strong base of known concentration until the pH rises sharply. Conversely, if the pH is above seven, then the unknown solution is basic, and so the titration will involve adding a strong acid of known concentration until the pH decreases dramatically.
Every time a small amount of titrant is added to the analyte, we measure the pH. If we plot the pH of the solution every time we add some volume of titrant, the result is a very useful graph - the titration curve. There are two key points on the graph of the titration curve that provide key information about the analyte. First, the equivalence point of a titration is the point where the analyte has been completely neutralized by the titrant (recall that the general form of a neutralization reaction Acid + Base → Salt + Water + Heat. The endpoint of a titration is the point where moles of acid equals moles of base. In our video on neutralization, we discussed how the type of salt formed dictates the pH of the solution at the equivalence point of neutralization. Therefore, the pH of the solution at the equivalence point is determined by the type of salt formed. For instance, a weakly acidic salt would produce a weakly acidic pH (less than 7). On the other hand, a titration between a strong acid and base would result in a salt with no acid-base activity, and so the pH of the resultant solution would be equal to 7. The equivalence point is the point on the titration curve where the slope is steepest. Looking at a titration curve, you’ll notice that the slope becomes almost vertical in its center. The very middle of this steep portion of the curve is the equivalence point of the reaction. Adding any extra acid or base to the solution at this point will most dramatically change the pH of the solution, because the acid and base involved in buffering have been neutralized.
Second, the half equivalence is the point on the graph where half of the analyte has been titrated. For weak acids and bases, a shortcut we can use is to say that solution pH = pKa of the analyte at the half equivalence point. Recall the Henderson-Hasselbalch equation, which describes the pH of a buffer system composed of some conjugate acid-base pair. With this equation we see that pH is equal to pKa only when log([A-]/[HA]) is equal to 0. This expression is equal to 0 only when [A-] = [HA]. Mathematically, we can see that the half equivalence point must be the part of the titration curve where the amount of acid and base in solution are equal.
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Пікірлер: 2

  • @asimgodil8008
    @asimgodil8008 Жыл бұрын

    for the example problem why is the volume of the base 50mL, how do we know to use the volume at the equivalence point?

  • @coco.chrisps

    @coco.chrisps

    9 ай бұрын

    at the EP, the solution is neutralized (the amount of added titrant neutralized the unknown sol'n). If it took 50 mL of 1 M NaOH to neutralize the reaction, then that must be proportional to the amount of the unknown volume and concentration of the analyte (that's why we use (na)[A](Va)=(nb)[B](Vb); this equation tells us that the amount needed to neutralize a reaction).