Diprotic & Polyprotic Buffers

by Jules Bruno
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Okay, calculate the pH of 75 MLS of 750.10 Mueller Phosphorus Acid Solution When ADM Elza 0.15 Mueller sodium hydroxide are added here, were given two K s. Now, before we approach this question, realize here that false force acid, which is H three p 03 looks like it will be a try product acid, where it would have three acidic hydrogen. But based on its lewis dot structure, it only actually has to que values. So phosphoric acid is not a try protic acid. It's a die product acid. And remember, when it comes to doing these types of questions, we have to analyze the information given to determine which form of the die protic acid is involved in the reaction. So we have the acidic form of phosphorus acid. When it loses its first H plus, it becomes H two peel three minus. And then finally, when it loses its last acidic hydrogen, it becomes HP all three to minus. So these are the three forms of the dive protic acid, and we have to determine which one of the of the forms are involved in this question. Now, remember, we have a strong base reacting with our weak acid here, and we set up above that. If we're dealing with these two forms, this would happen before the first equivalents volume. And then we'd use these two after the first equivalents volume. So here to determine that we say M acid times v acid equals M base times V base. All right, so we're gonna use the information of the week species, which is the acid. It has 0.10 Mueller and then it's 75 MLS. We know the molar ity of the Strong's bases. 15 Mueller. Now, remember, when we're looking for the equivalent volume, we're looking for the equivalent volume of the strong species, which in this case represents a strong base. We're gonna divide both sides now my 0.15 Mueller so V base equals 50 amounts. So that means we need 50 miles to get to the first equivalents point. And to get to the second equivalents point, we would need double that. We need 100. Immelt's. So we have 80 mls. So we've gone beyond the first equivalents point. So that means we're dealing with these two, and we're not above 100 MLS. So we're not dealing with stuff beyond the second equivalents point. Okay, So if we're dealing with stuff beyond the second equivalents point, um, that means that we'd really be looking at this form here, reacting with the base in some additional way. All right, so we know that we're in between basically the first equivalent volume in the second one. So we've gone beyond the first equivalent volume. So we're not gonna use these two. We're gonna be using these two because we've gone after the first equivalent volume, but before the second. So we're gonna have hte to peel three minus, which is the acid form. It's going to react with my strong base. So we have initial change final. So here. This is my acid form in this case or intermediate form, which is acting like an acid. This is my base. Remember, acids donate h plus, so the age plus will connect with the O. H minus of the base to give us water. What's left behind will be n a to p 03 h p Okay, so that's our structure. And we know that's the formula for for this product here because remember when this gives away an H plus, It becomes h p +03 to minus. It's gonna react with the sodium here, which is plus one. So the two comes here and the one comes here. So that's how I came up with this formula here, remember, In an I C F chart, we only care about three things we only care about. Whatever strong. We only care about the weak acid form, and we only care about its conjugate base form the fourth species we ignore. Next, we're gonna say our units have to be in moles, so divide the MLS by 1000 and multiplied by the mole Aridjis. When we do that, we get 0075 moles and 0.45 moles here. This is gonna be equal to zero because we're not talking about it all yet. And remember, the smaller moles will subtract from the larger moles. Yeah, so this is gonna be 0.30 moles. This is gonna be zero. Whatever happens on the product on the reacting side, opposite happens on the product side. So we're gonna add this many moles. So we're gonna say at the end. What do we have left? We have weak acid left. We have conjugate base left. So we have a buffer. So we use the Henderson hassle back equation. Remember, because we're dealing with these two forms here. They're connected to each other by K two, which means we're gonna use P k to within my Henderson Hasselbach equation. So p k two plus the log of conjugate base over the weak acid form. So p K two is just the negative log of K two, plus the log of the moles of the contract based over the weak acid. So when we do that, so take the negative log of this. Take the log of this expression, add those two numbers together. So here this would be 6.698 and this would be 0.176091 So when we add them together, pH equals 6.88 or so when we round. So this would represent the pH of our buffer system that we created again. Die poetic and Polly product buffers are tricky. You have to look to see which form you're dealing with when approaching any of these questions. That's the key to getting the correct answer