Our listener’s choice recaps three topics: (R)ICE tables, titrations and buffers.
Our listener’s choice recaps three topics: (R)ICE tables, titrations and buffers. We start with the RICE table - what does it actually stand for? (0:49). RICE tables are best explained with an example. In our case: The reaction of gaseous hydrogen with gaseous chlorine to form gaseous hydrogen chloride - a reaction that has a K = 49. And we are calculating the concentrations at equilibrium (1:44)! Two important tips for RICE tables: (1) practice, practice, practice; (2) make assumptions (4:09).
Our next two topics are connected: titrations and buffers. We start with the titration of a strong acid/strong base and discuss the calculation of the pH at several points along the titration curve (6:01). Our second example is the titration of a weak acid with a strong base. Again, we discuss the pH calculations along the titration curve, but then focus on buffers, which are formed when the weak acid is partially neutralized (7:29). The episode defines buffers and describes how they work (8:00). Moving along the titration curve, we discuss the calculations for buffers and the midway point of the titration (9:18), the calculations at equivalence point (10:03) and beyond the equivalence point.
Question: What will be the approximate pH of an equimolar solution of NH3 and HCl?
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Hi and welcome to the APsolute Recap: Chemistry Edition. Today’s episode is our third listener’s choice episode.
Let’s Zoom out:
It is crunch time! We are getting closer to the AP Exam and we’ve heard our listeners! ICE table, titrations and buffers are three of the more advanced concepts in AP Chemistry - at the same time, they are REALLY important. So let’s recap them together!
Let’s Zoom in:
Our first recap topic comes to us from listener Tiffany J. asking us about ICE tables. Depending on if you use the “R”, which stands for the balanced chemical reaction, they can also be called RICE tables. They are a SUPER helpful tool when it comes to equilibrium calculations, which are a key skill in units 7 and 8 - a great question! So let’s take a closer look: R stands for reaction, I stands for “initial concentration”, which can be in moles but is more commonly used with molarity. C stands for “change” and indicates, using the stoichiometric coefficients how much a chemical species will change during the reaction. E stands for “equilibrium concentration” and is therefore the concentration at equilibrium.
Let’s make an example from Unit 7 where we have to calculate the equilibrium concentration with a given K: gaseous hydrogen and gaseous chlorine react to form gaseous hydrogen chloride. The equilibrium constant for this reaction is 49. The balanced chemical equation shows: 1 H2 + 1 Cl2 yields 2 HCl. That is a key part. It means that my concentrations of H2 and Cl2 will decrease by factor 1, my concentration of HCl will increase by factor 2. But one step after the other. Let’s say we start with 0.5 M of hydrogen and 0.5 M of chlorine and enter them in our “Initial” row. The initial concentration of HCl is 0. For the change we use our stoichiometry coefficients: hydrogen and chlorine will decrease by x, whereas HCl will increase by 2x. For our equilibrium concentration that means hydrogen and chlorine are both 0.5-x and hydrogen chloride is 2x. Now we can set up our equilibrium constant expression. If you need a recap, listen to episode 25! K equals concentration of HCl squared over the concentration of hydrogen times the concentration of chlorine. Then, we plug in our “equilibrium concentrations” with their dependency on x. Additionally, we know that K is 49. Careful: The concentration of HCl is 2x and we have to square that according to the equilibrium expression. Students often get confused because the coefficient 2 shows up twice: as change and superscript! Now, this is an easier one to solve: We can take the square root and then solve for x, which is 0.389. Therefore the equilibrium concentration of HCl will be 2 times 0.389, which is 0.778 M. If you want to double check your answer you can plug the equilibrium concentrations into the equilibrium expression and you’ll get 49!
There are A LOT of variations to ICE tables, for example: moles given instead of molarities; calculation of the equilibrium constant, taking into account Q, and and and… they definitely need practice. One important hint: Sometimes you encounter problems where you would have to use a quadratic equation to solve for x. In these problems you will usually have a very small K, which indicates that the reaction heavily favors the reactants. This allows us to “make an assumption” that the x is so small it won’t significantly decrease the amount of reactant and therefore, we can ignore it in the sum or difference of the denominator. That will allow us to solve for x without a quadratic equation.
I would like to combine the second and third topic request coming to us from listeners Natalie B. and Eric H. - titrations and buffers. Titrations are a laboratory method that is used to determine the concentration of an unknown acid or base by adding a base or acid of known concentration. You can determine the concentration of the unknown, by titration to the equivalence point, which is defined as the point in the titration where the amount of moles in the analyte - that the unknown one - and the moles you added via titrant, are equal. Let’s talk about an example: Titrating a strong acid of unknown concentration with a 0.1 M solution of sodium hydroxide. Looking at the net ionic equation, we can see that the hydronium reacts with hydroxide to form water. At the beginning of your titration, your hydronium will be in excess and, until you reach the equivalence point, determine the pH. For these sections of the titration curve, we can use the concept of limiting reactant and excess reactant to determine the pH. At equivalence point, the moles of hydronium that are in the unknown acid and the moles of hydroxide added are equal. For our example, this would be at pH=7 because then all we have is water! If we keep titrating, the hydroxide is in excess and the pH is determined by the excess of hydroxide.
Now, what if I am titrating a weak acid with a strong base? Let’s talk about different “points” of the titration curve. The initial pH of the weak acid before any hydroxide is added can be calculated using the ICE table and the Ka value. Then we start adding hydroxide, which will react with our weak acid. The weak acid, however, is in excess and therefore you will have weak acid as well as its conjugate base in the beaker - and that’s a buffer! Wohoo!
Buffer solutions resist changes in pH when a small amount of strong acid or base is added. How do they do that? Their superpower comes from having large, but approximately equal amounts of weak acid and its conjugate base. When adding a small amount of strong acid, H+, to the buffer solution the H+ will react with the conjugate base, F- to form the weak acid, HF. This reaction lessens the impact on pH, because, even though we are increasing the concentration of HF and decreasing the concentration of F-, it doesn't really change the concentration of H+ - or H3O+ ions - in the solution. If the conjugate base wasn’t there and wouldn’t be able to react with the added H+, then the pH would change much more, because then you would directly increase the concentration of H+ ions and therefore lower the pH. When adding a strong base to a buffer, the OH- ions will react with the weak acid and form water plus the conjugate base. Now the weak acid is sacrificing itself, neutralizing the added OH- and therefore not significantly increasing the pH.
I’ve mentioned that the concentrations are approximately equal. If they are exactly equal, which corresponds to the point in our titration when you have converted half of the original moles of weak acid to conjugate base. This point is called the midway-point or half-equivalence point. Because the concentration of conjugate base and weak acid are equal and the pH = pKa. This is coming from the Henderson - Hasselbalch equation with states that the pH = pKa + logarithm of the concentration of conjugate base over acid. We can use the Henderson - Hasselbalch equation to calculate the pH at any point on our way to the equivalence point.
At the equivalence point, we have converted all our weak acid to conjugate base. The pH is therefore determined by the reverse reaction: the reaction of conjugate base with water to form the weak acid and hydroxide. And so, the equivalence point will be slightly basic. You can calculate the pH of the equivalence point by calculating the pH of the weak base - just be careful to use the Kb instead of the Ka and to have in mind that you are calculating pOH first! Beyond the equivalence point, the hydroxide is in excess and the pH is determined by the excess hydroxide.
My two examples obviously described the titrations of a strong and weak acid with a strong base. You can, of course, also do it the other way round and have a strong or weak base of unknown concentration and titrate it with a strong acid. The calculations are, as we say, same, same, but different.
To recap:
ICE or RICE tables can be used to calculate the equilibrium concentrations or the equilibrium constant. They take into account the stoichiometric coefficients, which indicate the “Change”. Titrations are a laboratory method that is used to determine the concentration of an unknown acid or base by adding a base or acid of known concentration and titrating to the equivalence point where the moles of analyte equals the moles of titrant. The basis for the calculations for titrations of strong acids/bases is the stoichiometric concept of excess species. The equivalence point for a strong/strong titration is at pH=7. When titrating a weak acid/base with a strong base/acid, we are creating a buffer. Buffer solutions resist changes in pH when a small amount of strong acid or base is added, because they have large, but approximately equal amounts of a conjugate acid-base pair.
Coming up next on the APsolute RecAP Chemistry Edition: Multiple Choice Strategies.
Today’s Question of the day is about titrations.
Question: What will be the approximate pH of an equimolar solution of NH3 and HCl?