You are asked to prepare a pH = 3.00 buffer solution starting
fr...

Question

You are asked to prepare a pH = 3.00 buffer solution starting 
from 1.25 L of a 1.00 M solution of hydrofluoric acid 
(HF) and any amount you need of sodium fluoride (NaF). 
(b) How many grams of sodium 
fluoride should be added to prepare the buffer solution? 
Neglect the small volume change that occurs when the sodium 
fluoride is added.

Accepted Answer

Welcome back everyone in this example, we're considering a 250 mL solution of 2500.20 moller propane OIC acid. We're going to calculate the mass of sodium propane it needed to produce a buffer with a ph of 5.11 assuming that our volume of our solution will not change. So what we should recognize is that propane OIC acid is not on our list of memorized strong acids. So it's a weak acid and our salt mentioned sodium propane. It we we should recognize as assault because we want to recall that to produce a buffer. We're going to take our weak acid and mix it in a solution with its salt where we recall that the point of a buffer is to resist change in ph And so that's why according to the prompt, the ph is going to stay at 5.11. So it's important that we write out our dissociation equation of our weak acid when it's in solution. So we should recall the formula for propane OIC acid, which is going to be C3H6 and then 02. This is soluble in water. So we're going to dissolve it in liquid water and this is going to be an equilibrium with its conjugate base, which we should recall is going to be our propane OIC acid losing a hydrogen atom to form a bond with water so that we produce our conjugate base being C three H 502 with a minus one charge since we lost a proton and then we form hydro ni um as well as a product. So these both get acquis labels since they're ions because this is a H 30. Plus for our hydro ni. Um So just to label everything, this is our propane OIC acid which is our weak acid. And then we have our conjugate base which is our propane away ion. And we can say it's an an eye on since it has a negative charge. So now that we understand the reaction happening, we want to recall the formula where we take our ph which is calculated from taking our P. K. A. Of our weak acid added to the log of our quotient where we take our concentration of our conjugate base. So this should be in brackets divided by our concentration of our weak acid. So this completes our formula and we want to plug in what we know now we know that our propane OIC acid is our weak acid. And if we look up online or in our textbooks we would see that it has a P. K. A equal to a value of 4.87. So we're going to plug that into our formula. So we can say that our ph which given in the prompt is a value of 5.11. So we'll plug that in as 5.11 is equal to R. P. K. A. Which we determined for propane OIC acid is 4. added to the log of our quotient where we have our concentration of our conjugate base which is our prop annoyed and ion. Now we don't know that concentration is not given to us in the prompt. So we're just going to write that in as C three H 502 minus Divided by our concentration of our weak acid which given from the prompt is a value of 0. moller for propane OIC acid. So what we want to do is simplify here. So we're going to subtract 4.87 from both sides. That's going to cancel out on the right hand side and then on our next line we'll be able to simplify to the difference of our log of our concentration of our conjugate base prop annoyed an ion Divided by our concentration of our weak acid, phenolic acid .20 moller. This is set equal to our difference of our left hand side which is a value of 0.24. So scrolling down for more room, we want to continue to simplify by canceling out that log term. So we're going to go ahead and recall that to cancel out the log term. We make our left hand side and exponents to the base of 10. And so the right hand side is also going to be an exponents to the base 10. So that's going to cancel our log term. And what we're going to get is now will simplify to our concentration of our conjugate base prop annoyed and ion divided by our concentration of our weak acid propane, OIC acid 0.20 Moeller equal to the right hand side which gives us the result of a value 1.74 when we round up. So what we're solving for here is our concentration of our conjugate base prop annoyed and ion. And so because we have a diagonal here we can go ahead And switch the place of our concentration of our conjugate acid with the 1.74 value. And we will be able to simplify to our concentration of our conjugate base where we would say that our concentration of prop annoyed and ion Is equal to a value of 1.74 divided by 0.20 moller. And that's going to give us a value of 0.348 moller. Now we want to recall that our term polarity is equivalent to units of moles divided by leaders. So we can interpret this as . moles divided by leaders of our solution as our concentration of our conjugate base. Now this is important because we need to ultimately enter this question. A unit being mass of our salt, sodium propane. It. So using our interpretation of our polarity in units of moles per liter, we can go from moles and get two g using a dimensional analysis step. So what we want to first recall is our and actually let's just say that this is part one of our solution. And so now we're going to go to part two of our solution where we're going to make note of our molar mass of our salt which is our sodium propane. It. So that formula would just be let's change your pen. So C three H 502 N. A. This is our salt. And when we refer to our periodic table to find its molar mass, we see that it has a value equal to 96.07 g per mole according to our periodic table. So what we're going to do is use a dimensional analysis step where we're going to take that concentration of our conjugate base of our weak acid where we said that that's a value of .348 moles per liter. And we're going to multiply to cancel out our unit of leaders by utilizing our conversion factor to go from milliliters to leaders. Where we recall that our prefix milli tells us that we have 10 to the negative third power of our base unit leader to one mil leader. This allows us to cancel out our units of leaders center to line diagonally. And now we can focus on canceling out that unit middle leader. So we want to go back to our prompt and recognize the volume given for our solution and that according to the prompt is a ml solution that we have. So we're going to use that as a conversion factor and say. So if we go back to our equation, we want to verify that it's balanced. And actually to completely balance our equation out, we're going to place a coefficient of three in front of our water and a three in front of our hydro knee. Um now we should recognize that that volume given for our Solution is for one mole of our weak acid propane OIC acid based on our balanced coefficients. And so that's what we'll use as our as our conversion factor below. So we have in our numerator, 250 ml of our solution for one mole of our acid. We'll say weak acid. I'm sorry. Let's make this clear. So one more of our weak acid. Now this allows us to cancel our units of middle leaders and as we stated, we want grams to be our final unit. So we're going to incorporate that molar mass that we determined above for our our sodium salt. So we would say that we have a molar mass of 96 point oh seven g of our salt C three H 502 and a And this is for one mole from our periodic table molar mass. So now we can finally cancel out those units of moles, leaving us with grams as our final unit. And this is going to give us our mass of our salt needed to produce the buffer Equal to a value of 8.358g. And so this is actually going to be our final answer to complete this example, again as the mass of our salt needed for the buffer solution. And so to be extra clear, we want our minimum amount of sig figs, so that would be two sig figs. So our final answer is going to be 8.4 g. So what's highlighted in yellow here is actually going to be our final answer. To complete this example. I hope everything I reviewed was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.

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