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# Buffer and Acid/Base Chemistry CH 344

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This 4 page Class Notes was uploaded by Cody Brazel on Saturday August 29, 2015. The Class Notes belongs to CH 344 at Southeast Missouri State University taught by dr. Jacoby in Spring 2015. Since its upload, it has received 24 views. For similar materials see Biochemistry in Chemistry at Southeast Missouri State University.

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Date Created: 08/29/15

Notes For Biochemistry CH531UI331 Notes for 824828 2015 0 Nearly everything learned during our first week in biochemistry has been strictly review over the cellular structure acid base chemistry and thermodynamics Due to the extensive lecture he gave concerning cellular structure and the PowerPoint slides he has shared these notes will focus primarily around the intricacies of acid base chemistry The next set of notes will catch up on thermodynamics and the new material Dr Jacoby will cover in class How to Quantify Acids and Bases practice calculations will be given after the body of notes Equations to know 1 LogHpH 2 LogOH39pOH 3 10quotquotHH 4 1039p HOH39 5 14pHpOH 6 KWOH39H As Dr Jacoby has undoubtedly expressed the interactions between acids and bases permeates nearly every chemical reaction within organisms The key scale by which scientists measure the amount of acid and by extension base is pH H present within a given solution So algebraically the amount of Hydrogen Ions can be found by taking the inverse of the negative log which happens to be 104 In addition pOH is negative logarithm of the amount of Hydroxide ions found within a given solution and many of you may be wondering just how to find pOH Luckily a simple equation that links pH to pOH can be used By using this equation one can find the pOH from pH and then find the amount of Hydroxide ions within the solution Now Let s do a few different examples Questions and solutions 1 If a given solution has a Hydrogen ion concentration of 131O395 M what is the pH of the solution a Solution By taking the negative log of the hydrogen ion concentration the pH of the solution can be found Therefore Log1310395pH489 2 Using the same Hydrogen Ion Concentration from question one what is the concentration of hydroxide ions This can be solved in two different ways a One way of finding the answer would be to use the equations 15 and 4 Luckily we have already found the pH of the solution using equation 1 so the second step of this problem would be to rearrange equation 5 14pHpOH therefore pOH14489911 b The second way would be through the use of equation 6 The Kw otherwise known as the dissociation constant for water is equal to 1103914 M If we know the amount of Hydrogen ions present within the solution 1310395 we can rearrange equation 6 to solve for the hydroxide ion content OH39 110 141310395769103910 M 3 Find the pH the Hydroxide content and the hydronium content of a solution if you were given the pOH 10 a Let s make this a selflearning question Find each value through the use of the equations shown The beauty of acid base math is that one can start from almost any direction or equation Note that you do need to start out with an equation that uses pOH hint I would start with equation 5 to quickly find pH The HendersonHasselbach Equation and How to Make Buffers equation pHpKaLogA39HA Dr Jacoby has expressed a large amount of sentiment in terms of how to make buffers Formulating buffers and knowing how to mix them is an absolute essential for anyone wanting to pursue any level career in biology The method I use will focus around his teaching within the second video For clarification when I begin doing calculations I will use the exact same numbers and values he uses in his buffer calculations of video 23 on moodle The method I use and the method the professor uses are slightly different BUT NEITHER method is better than the other If you find that the method I use is uncomfortable for you please use Dr Jacoby s method for it will ensure you get the greatest amount of points First and foremost one must have a good understanding of the Henderson Hasselbach equation in order to make a buffer Buffers are a mixture of a weak acid and the weak acid s conjugate base Conjugates are similar species chemicals that usually defer by one hydrogen For example an acid s conjugate base has one less hydrogen and vice versa The Henderson Hasselbach equation links the acids pKa the pH of the solution and the ratio of base to acid Dr Jacoby has also expressed that the way he has done it in class will be the way he wants us to solve a similar question for test In short if you master this method or the method he uses in the video you WILL get full points on the question Here we go THE METHOD The question wants to know how much acid and how much base in grams we need to add in a solution of water to attain a certain pH What has been given 1 The pH of the solution should equal 70 We are using a phosphate buffer so in short we are using various forms of phosphoric acid The Molarity of the solution is 015 MolesLiter The total volume of the solution is 3 Liters The form of phospohoric acid that has a pKa closest to the desired pH is monosodium phosphate pKa686 91quot The first step to success is to write out the HH equation pHpKaLogA39HA Since we know the pH and the pKa of the acid we are using we can plug those values into the HH equation It may look funny at first but we are trying to find the ratio of A to HA So let s make A39HA equal to x By plugging in these values and variables the equation will look like this so Using algebra knowledge we should attempt to isolate x The first step is the subtract 686 from both sides giving this equation 14ogx The inverse of a log is 101 so raise each side to the power of ten 103914x138 138 is the ratio of the conjugate base to the acid Therefore we can show the relationship as a ratio like so 138A 1HA Here is where things may start to get a little confusing We know that we have 138M of base for every 1M of HA so the total amount of Molarity shown here would be equal to 1381238 It should be inferred that the acid and base are occupying different fractions By finding the exact fraction of the solution the acid and the base take up we can figure out the specific MolesLiter of the acid and the base Here is the mathematical interpretation of what I am talking about If there are 138 M of the base with at total 238M the fraction of the solution the base resides in is equal to 138238 58 Let s do the same for the acid If there is 1M out of a total 238M the fraction is expressed as 123842 Now that we know the percentile each acid and base takes up in a solution let s find out how much acid and base would be present in a solution of 015M For each scenario times the percent to the solutions molarity The molarity of acid present in this solution will be equal to O150420063M The molarity of base present in the solution will be equal to 0150580087M Now we are almost done just a couple more concepts and we are home free Now that we know that molarity of each acid and base we can find the exact amount of moles of the acid and base Since Molarity is equal to MolesLiter if we multiply each molarity by the volume of the solution 3 liters the Liters will cancel out and leave us with moles To quantify this here are the steps The moles of acid is equal to the molarity we found times 3L 0063M3L0189 Moles The moles of base is equal to the molarity we found times 3L 0087M3LO261 Moles Going back to CH185 we can find the grams needed of each acid and base by multiplying the grams found by their molar masses This is because molar mass is equal to GramsMoles so by multiplying the Moles found by the molar mass we are left with just grams Early on we determined that the acid was monosodium phosphate which has a molar mass of 11998 GramsMole Since the conjugate base has one less hydrogen than the acid it is inferred that the base is disodium phosphate with a molar mass 14196 To determine the amount of grams needed for the base multiply the moles found by its molar mass 0261 Moles 14196 Gramsmole371 grams of Disodium phosphate To determine the amount of grams needed for the acid multiply the moles found by its molar mass O189Moes11998 GramsMole227 Grams of monosodium Phosphate This concludes the notes over acid base chemistry I highly recommend googling around for acid base questions in order to test your newfound knowledge For reference the method of buffer calculations used were derived from searching for various practice problems I hope this helps and if you have any questions please contact me at cibrazdazgmailcom

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