Suppose you want to do a physiological experiment that calls for a pH 6.50 buffer. You find that the organism with which you are working is not sensitive to the weak acid H2A 1Ka1 = 2 * 10-2; Ka2 = 5.0 * 10-72 or its sodium salts. You have available a 1.0 M solution of this acid and a 1.0 M solution of NaOH. How much of the NaOH solution should be added to 1.0 L of the acid to give a buffer at pH 6.50? (Ignore any volume change.)

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Hello everyone today. We have the following problem. In a laboratory experiment we need to prepare a solution buffered at ph equal to six. The organism that you're working with is not sensitive to the weak acid moronic acid which has the following K. A. Values. And it's sodium salts, solutions of 1.5 moller moronic acid and 1.5 molar of sodium hydroxide are available in the laboratory, calculate the volume of sodium hydroxide solution will be at that should be added to the 1. liters of melodic acid. In order to create a buffer solution buffered at ph of six, assume that volume does not change. So the first thing that we have to do is we have to calculate our concentration of hydro ni um units with that ph. And so what we can do is we can do 10 to the negative of the ph Which we plug that in. That's going to be to the -6. And that gives us a value of 1.0 times 10 to the - molar. So now we have our concentration of hydro ni um ions. Next we need to calculate our ratio of our like acid or the eye on that it produces to the actual acid. So that is going to be equal to are we going to use Our 2nd K. A. value to do so. And so we can say that our second K. A value is equal to this metallic acid ion times the hydrogen concentration over maloney asset. Furthermore, we can rearrange this too, that our second K. A value is going to be divided by our hydro knee um concentration and that is going to be equal to our two times 10 to the negative six Over our one times 10 to the -6. This is going to give us a value of two. So since both components are in the same solution or the same volume, the ratio can also be in terms of moles. So we can say that we have two moles or just two in general. So we can say that we have two moles of our moronic acid. So in the 1.5 liter solution of this moronic acid tube and then we have our 2.25 moles of the material that contains the ion. And so therefore the moles of our melodic acid Can be added to the two moles Or can be added to two moles times the moles of our moronic acid ion. That's gonna give us 2.25 moles. We're gonna say that that equals 2.25 moles. And then We're going to place a three malls of our ion Is you go to 2.25 giving us a total moles for our moronic acid ion To be equal to 0.75 moles. Next we need to find the number of moles of our secondary ion that forms when we remove another proton. That's going to stem from the 2.25 moles subtracted by the 0. moles we just sold for to give us 1.5 moles. And so next we have to calculate the amount of that 1.5 sodium hydroxide that has to be added to produce the necessary moles. And so we're going to do with stepwise neutralization. So we're gonna have to I. C. F. Tables. So for our first I cf table, we're going to have the first process happen. Where are moronic acid? There reacts with our sodium hydroxide to form this product and water Initial change in our final concentrations. We started with 2.25 moles of our moronic acid and our base as well. We started with zero for our product. And of course water is not going to be included in this. And we are we are tight trading with our sodium hydroxide. So we're going to subtract by that amount 2.25. And then we're gonna add that amount to our products for our secondary I cf table. So this is our I. C. F. Table one, B R I C. F. Table two, we're going to set up our secondary process reacting with another sodium hydroxide, teal, us that final ion and water. And once again water is not included, We have our I. c. and F. We started with 2.25 moles of our equivalent. We don't know what our sodium hydroxide is. And of course we had zero for our I on. So when we subtract by X. Which is our which is our tight Trent. And then we add that to our products. We're going to get a final concentration of 0.75 moles, zero moles for the sodium hydroxide and 1.5 moles for our ion. And so the total moles the normals Can be calculated of sodium hydroxide. Of course, by the 2.25 moles that we started with plus the 1.5 that we added to give us 3.75 moles. And so lastly to find our volume of our sodium hydroxide, We can take that 3.75 moles and multiplied by the conversion factor. Or our molar itty, which is our one leader is over 1.5 moles To give us 2.5 L of sodium hydroxide. Of our answer. Overall, I hope this helped. And until next time

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Key Concepts

Buffer Solutions

Henderson-Hasselbalch Equation

Acid-Base Neutralization