Chem 330 week 2 notes
Chem 330 week 2 notes Chem 330
Popular in Quantitative Analysis Chemistry
Popular in Chemistry
This 3 page Class Notes was uploaded by Leslie Pike on Wednesday June 15, 2016. The Class Notes belongs to Chem 330 at Western Kentucky University taught by Dr. Darwin Dahl in Fall 2015. Since its upload, it has received 19 views. For similar materials see Quantitative Analysis Chemistry in Chemistry at Western Kentucky University.
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Date Created: 06/15/16
Acids and Bases, advanced You can’t take quantitative analysis chemistry without having taken intro chemistry, so I’m assuming that you’ve been introduced to the concept of acids and bases and pH. You were probably taught to calculate the pH of an acidic solution as –log(acid concentration). This works just fine, until we make up a solution of 10^-8 M HCl and you are asked to calculate the pH. –log(10^-8) = 8, but a pH of 8 means that you have a basic solution. This is obviously not the case, seeing as your solution consists of water and hydrochloric acid! Where did we go wrong? Water can auto-ionize, K=10^-14. As you can see from the very small K value, the auto-ionization of water has a very slight effect on pH (especially if there are other, stronger, acids and bases in solution), and in freshman chemistry you were taught to ignore it. However, in very dilute acidic and basic solutions, the auto-ionization of water makes an appreciable impact on pH. Therefore, we must take it into account in our pH calculation. First, we write our mass balance equation. In this case, hydrogen ion comes from two sources, dissociation of hydrochloric acid and auto-ionization of water. [H+]=[Cl-]+[OH-] The Cl- concentration is known, because it is equal to the HCl concentration. Using the Kw expression, we can remove the variable OH- from the equation: [H+]=10^-8 + Kw/[H+] This is then solved with the quadratic equation. We find that the pH is 6.98. This is a reasonable answer, seeing that the pH of pure water is 7, and this solution is known to be very slightly acidic. Preparing a Gran plot from a titration The convenient thing about a Gran plot is that it is linear (unlike the typical titration curve plot), enabling us to easily fit an equation to the data gathered in lab. When you are titrating an acid with a base, you use this equation: Vb[H+] = -Ka*Vb + Ka*Veq [H+], dependent on Vb (the volume of basic titrant you have added, a variable), is the “y” of your y=mx+b equation. Vb is your “x”. All other values are constants. Your linear fit to your data will be a line with a negative slope; the point at which the line crosses the x-axis is your Veq, or equivalence point volume. Ka can be determined from a gran plot as: -1*slope of fit line. Alpha fractions Alpha fractions represent how much of a weak acid is present in the protonated form, and how much is present in the deprotonated form. α0is the amount of fully protonated acid (0 protons have been lost) divided by the total amount of acid, α 1epresents the amount of acid that has lost 1 proton (also the Ka 1cid) divided by the total amount of acid. If the acid is polyprotic, α 2s the amount of acid that has lost 2 protons (also the Ka acid) divided by the total 2 amount of acid, and so on. It is easy to get this numbering backwards, so remember that the subscript number represents the number of protons that have been lost. The sum of all the alpha fractions for any given acid is 1. We need know only the dissociation constants and the pH to calculate alpha fractions. The denominator is calculated as follows: (# of protons acid has) (# of protons -1) (# of protons-2) [H+] + [H+] Ka 1 [H+] Ka 1a +2… Continue in this fashion until you have accounted for all the protons your acid has. Your final term will consist of no [H+] and will contain all of the Ka’s. The numerator will be one of the terms in the denominator equation. For α , it is the 0 first term. For α ,1it is the second term, etc. Sample problem: we have 0.1 M phosphoric acid, H PO . This is3a w4ak acid. The 2- solution pH is 4. What fraction of the acid is present as HPO ? What is4the concentration of HPO ? 2- 4 We want the alpha expression for the acid that has lost two protons: α = [H+]Ka Ka /([H+] + [H+] Ka + [H+]Ka Ka + Ka Ka Ka ) 2 1 2 1 1 2 1 2 3 We are given a pH of 4, so [H+]=10^-4. The Ka’s we would be given, or we could look up in a table. We could find the concentration of the acid missing two protons by multiplying its alpha fraction by the total acid concentration. Titration of a Dilute Polyprotic Weak Acid You titrate 50 mL 0.05 M H A wit2 NaOH. K1=0.021, K2=6*10^-6. What does the resulting titration curve look like? Calculate the pH when 0, 10, 25, 40, 50, and 60 mL of NaOH have been added. If K1/K2 > 1000, you will see two distinct breaks in the titration curve, instead of just one break like you see for a monoprotic acid. This is the case in this situation. For this problem, our weak acid is actually quite strong, and it is also dilute. We CANNOT use simplifications because of this; we have to work everything the long way. We will use K1 for our Ka until we have used up all of the H A, then we2will use K2. (Makes sense because K1 is the dissociation constant for H A and K2 is 2he dissociation constant for HA-.) 2 At 0 mL NaOH added, we have no base in solution. Ka=x /([H A]-x), solving2for x and taking the negative log of this gives pH 1.63. At 10 mL NaOH added, we use ICE and find that we have 1.5 mmol H A and 1 mmol 2 HA in solution. We DO NOT use Henderson-Hasselbalch equation, because our Ka is too high and our concentration is too low. We use the unsimplified version: Ka=[H+]([HA-]+[H+])/([H A]-[H+]2 Plugging in numbers and solving gives [H+]=0.0108 and a pH of 1.97. At 25 mL, the OH- has converted all H A to HA2. We only have HA- in solution, but HA- is an amphiprotic material. How do we tell what it does? We need a new equation for this (Dahl refers to it as “big” because it is pretty big). [H+]=sqrt(K2*Ca/(1+Ca/K1)) Plugging in our values gives [H+]=0.000279 M and pH=3.55. Now that our H A i2 gone, we will be using K2 for our Ka value. K2 is small enough that we can use our regular equations to solve for pH, instead of the unsimplified ones. At 40 mL, do an ICE table, and you will get 1 mmol HA and 1.5 mmol A . Use 2- Henderson-Hasselbalch. Use K2 for your Ka. pH=pKa+log(A/HA)=5.22+log(1.5/1)=5.4 At 50 mL, you only have A in solution. Use [OH-]=sqrt(KbCb). Your base, A , 2- contains no protons, so you will use Kb . Ka K1 =Ka1Kb 2Kw, s2 fo1 Kb we 1 substitute Kw/Ka . Don’t forget to adjust the concentration (it is now 0.025 M 2 instead of 0.05 M) to make up for the volume added during the titration. [OH-]=6.45*10^-6, pH=8.8 2- At 60 mL, you have A and OH- in solution. The OH- is a much stronger base, so we do ICE to calculate the amount of OH- and then calculate pH=14+log[OH-]=11.96. Mixture of Strong and Weak Acids and Bases You mix 20 mL 0.2 M HCl, 5 mL 0.1 M Ba(OH) , 20 mL 0.2 M H CO , and 5 mL 2.1 3 Na 2O . 3hat is the resulting pH? First, convert everything to mmol. (For the sake of simplicity, we represent carbonate as A .) We have 4 mmol H+, 1 mmol OH- (not 0.5 mmol, pay attention to 2- the formula), 4 mmol H A, an2 0.5 mmol A . From a kinetics perspective, the strong and weak acids are going to react first. 1 mmol of H+ will react with the 1 mmol OH- to give 1 mmol water. In solution, we now have 3 mmol H+, 4 mmol H A, and 0.52mmol A . 2- We still have strong acid left. It will react with the remaining base. 0.5 mmol of H+ will react with 0.5 mmol A to form 0.5 mmol HA-. In solution, we now have 0.5 mmol HA-, 2.5 mmol H+, and 4 mmol H A. 2 We still have strong acid left. HA- is amphiprotic, but the strong acid forces it to act as a base. 0.5 mmol H+ will react with the 0.5 mmol HA- and convert it to 0.5 mmol H 2. In solution, we now have 2 mmol H+ and 4.5 mmol H A. 2 We don’t have any bases left, only acids, so the reaction is done. Ignore the pH contribution of the weak acid. The pH is calculated as pH=-log[H+]=-log[2/50]=1.4. (We get the 50 from adding up the volumes of the various components of the mixture.)
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