A 1.0-L buffer solution initially contains 0.25 mol of NH₃ and 0.25 mol of NH₄Cl. In order to adjust the buffer pH to 8.75, should you add NaOH or HCl to the buffer mixture? What mass of the correct reagent should you add?

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Welcome back everyone identify whether to add potassium hydroxide or hydroponic acid to the buffer mixture. To adjust the ph to 9.56. The buffer solution of one leader initially contains 10.5 molar in methyl amine and 0.5 moller in methyl ammonium bromide, calculate the mass of the correct reagent needed. We're going to begin by recognizing that methyl amine is a weak base and our methyl ammonium bromide is a salt of our weak acid, methyl ammonium. And we want to also make note of our conjugate acid of methyl amine, which would be when methyl amine accepts a proton, which we would form C H three N H three plus. And so with this in mind now we want to recognize that since we're using this week base, we're going to want to find the ph of our initial solution by referring to our formula for P O. H. Which relates our PKB of our weak base to the log of the ratio between our conjugate acid over our concentration of our weak base methyl amine. So let's just put that in as weak base. So let's plug in our knowns But we need to also make note of our pKB of methyl amine which in our textbooks or online will find a value equal to 3.35. And so plugging in our knowns we have P O. H equal to 3.35 plus. The log of our concentration of our conjugate acid which is going to be the same as our concentration of methyl amine given in the prompt as 0.500 moller for our conjugate acid here And then in the denominator. We have our concentration of our conjugate base given as also .500 moller for methyl amine in the prompt. And so because we see that these values are the same, we'll just have the log of one. So in the next line we have our P O H equal to 3.35 plus the log of one. And we'll see that the log of one is equal to zero, meaning that our P O H is equal to 3.35. And now we want to recall that to get our p P H. We have our P O. H added to our ph which equals the value of 14. So this is a formula for memory just like this one. And we would rearrange this so that we said that our ph is equal to 14 minus 3.35, which equals a ph of 10.65. So this is our initial ph of our solution. And now that we have this information, we want to find our ratio Of conjugate acid too weak base needed to adjust the ph to 9.56. So we're going to use the ph that we just found to help us where or sorry not the ph we just found but we're going to use the PKB that we noted down for our weak base as well as the target ph given in the prompt. So this is to find the ratio and we'll use a different color for this part. So ratio to adjust Our Ph of Our Solution to 9.56. What we'll find is that we can use the formula for P O. H. Which we stated is equal to our value of minus R P H. Which in the prompt is a target ph of 9.56. And this gives us our P P O. H. Now equal to 4.444. Our final conditions after the buffers added. And so let's note that down once buffer is added now we just need to find that ratio between our conjugate acid and weak base for the ph to adjust to 9.56. So we're going to plug in our knowns, we know that our P O. H. Is now 4.441 step buffers added and this is equal to R P K. B of methyl amine 3.35 and added to the log of our concentration between are conjugate acid of methyl amine ch three NH three plus over the concentration of our weak base. Methyl amine C. H three and H two. And so simplifying this. We're going to subtract 3.35 from both sides Which is going to cancel this out on the right hand side and now will have a value of 1.09 equal to the log of our unknown ratio that we're solving for here. So a ratio between conjugate acid and weak base. And so to cancel out that log term, we're going to make both sides and explain it to 10 so that it cancels log out and what we'll have is that our ratio between our weak acid C. H three and H three plus and our weak base, methyl ammonium, Sorry methyl amine is equal to a value of 12.30-7. Now from the prompt we need to decide whether to add hydroponic acid as our buffer or adding potassium hydroxide. And recall that hydroponic acid is a strong acid and potassium hydroxide is a strong base. So in order to get to that Initial or sorry, in order to begin from our ph of 10.65 that we found to get to our target ph of 9.56. As the prompt mentions, we're going to need to lower the ph obviously. And so if we were to add a more basic species like potassium hydroxide that would increase our ph further closer to 14. But we need our ph to decrease to 9.56. So we're going to want to add a strong acid like hydrochloric acid to lower our ph. So let's circle that as our selection. So yes, to lower PH 9.56 and adding our hydroponic acid is also going to completely consume our weak base so fully consumes our methyl amine. We know that it's going to fully consume our methyl amine because adding a strong asset is going to increase our concentration of our acidic compounds. So our next step now that we know to add hydroponic acid is to figure out how many moles of this acid to add. So in order to figure out the moles of hydrochloric acid to add, we need to calculate our initial moles of our species. So we need our initial moles of our weak base and our salt. And from the prompt first finding our moles of our weak base methyl amine, We know that its concentration initially is .5 molar and we know that polarity Can be interpreted in terms of moles per leader. So we're going to write this as .5 moles per liter and then we're going to multiply this by the volume given in the prompt of our solution being one leader. And so this will leave us with units of moles. And what we'll find is that we have an initial mole of .5 moles for our week based methyl amine. And now for our moles of our salt, methyl ammonium bromide. We are given an initial concentration of .5 Molar, which we're interpreting as multiple leader and multiplying by the volume of the solution. One leader to cancel out our units of leaders leaving us with moles and we'll find that we have the same amount .5 moles. So, knowing that these are our initial moles of our solution, we can use this to relate it to the ratio between our conjugate acid and weak base. And use that ratio as well as our initial moles of each of our re agents to find the moles of hydrochloric acid to add. And so what I mean is we're going to relate our ratio with our conjugate acid moles initially plus the acid that will add X. So plus hydrochloric acid. Here we'll say is X. This is going to be I'm sorry. This should be parentheses and then brackets on the outside. This is going to be divided by our concentration where we have our weak base moles initially added to Our x moles of hydrochloric acid to add. And we're relating this to the ratio that we solved for between our conjugate acid and weak base equal 12.30-7. So, plugging in what we found initially for our initial moles will have .5 moles plus X divided by 0.5 moles plus X equal to 12.30 to 7. So sorry, just before I get too far ahead of myself, I just want to make a correction to our denominator. It should be minus X. Because as we stated if we're adding our acid, we're going to be consuming our base. So we should have minus whatever amount we are trying to solve for of our hydroponic acid, which is why we have minus X in the denominator. And that means that for our next step to simplify to sulfur X, we want to multiply both sides by 0.5 moles minus X so that it canceled out on the right hand side. And then in our next line we would have 0.5 moles plus X, isolated equal to 12.30 to times 70.5 moles minus X. And this is going to end, Sorry, we can just write out what that would distribute to. So just to show how that appears we're going to distribute so that we get 6.15135 moles minus 12.30 to 7 X. And then to simplify next, we're just going to say minus X on both sides. That will cancel it over here and we'll have 70.5 moles isolated Equal to 6.15135 moles. And then we're going to have minus 13.30 to X. So to continue isolating, we're going to subtract or yes, subtract 6.15135 moles on both sides. And this is going to simplify too Just X. Or sorry, negative 5.65135 moles equal to negative 13.30 to 7 X. Where we're going to divide both sides by -13.30-7. And so we would have x fully isolated to equal a value of 0. to 48 to seven moles. Which we can round to about 46 figs as 460.4 to 4 eight most. And so now we just need to convert this to gramps for our mass of hydroponic acid that will be adding to our solution. So we'll begin with 0.4248 moles of hydrochloric acid. And we're going to multiply by the molar mass of hydroponic acid which in our periodic tables will see or using our periodic tables will see. One mole is equal to 80.91 g of HBR And so canceling out molds. We're left with grams of HBR and this will yield a mass of 34 0.37 g. Which we can round to just 34 g as two sig figs. And this is our mass of our acid hydrochloric acid that will add to get to our target ph of 9.56 according to the prompt. I hope everything I reviewed was clear. If you have any questions, please leave them down below and I'll see everyone in the next practice video

A 1.0-L buffer solution initially contains 0.25 mol of NH₃ and 0.25 mol of NH₄Cl. In order to adjust the buffer pH to 8.75, should you add NaOH or HCl to the buffer mixture? What mass of the correct reagent should you add?

Verified Solution

Key Concepts

Buffer Solutions

Henderson-Hasselbalch Equation

pH Adjustment with Strong Acids and Bases

A 1.0-L buffer solution initially contains 0.25 mol of NH3 and 0.25 mol of NH4Cl. In order to adjust the buffer pH to 8.75, should you add NaOH or HCl to the buffer mixture? What mass of the correct reagent should you add?

Verified Solution

Key Concepts

Buffer Solutions

Henderson-Hasselbalch Equation

pH Adjustment with Strong Acids and Bases

A 1.0-L buffer solution initially contains 0.25 mol of NH₃ and 0.25 mol of NH₄Cl. In order to adjust the buffer pH to 8.75, should you add NaOH or HCl to the buffer mixture? What mass of the correct reagent should you add?