A 25.0-mL sample of 0.125 M pyridine is titrated with 0.100 M HCl. Calculate the pH at each volume of added acid: 0 mL, 10 mL, 20 mL, equivalence point, one-half equivalence point, 40 mL, 50 mL. Sketch the titration curve.

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Welcome back everyone. A 20 mL sample of 0.1 50 molar three methyl pine is titrated using a 0.1 25 molar hydro brom acid tyrant plot. The missing point corresponding to 30 mL of 0.1 25 molar hydro brom acid. And sketch the titration curve below below. We are given an incomplete titration curve with six points where we are missing one point for the PH recorded on the Y axis and the volume of tyrant recorded on the X axis and milliliters. So we're going to need to figure out what that missing point is. Our first step is to write out our reaction. We have three methyl Peridin represented by its formula C six H seven N reacting with hydro Bromma acid. So let's recall from our textbooks that hydroponic acid is a strong acid and three methyl Perine is not on our list of strong bases. So it is considered a weak base and thus will not fully dissociate. And so we would have our strong acid recall based on our Lewis definition act as a proton donor and donate a proton to our weak base, which will in turn form our first product, which is our conjugate acid of our weak base three methyl Perini CC six H seven N with an H now bonded and A plus one charge. So this is our first product. And as a result of our acid donating a proton, our second product is going to be our bromide one minus anion which will act as a spectator and will not influence. Ph take note that we have a 1 to 1 molar ratio between our weak base and strong acid. And recall that this means that at the equivalence point of our titration curve, the moles of our strong acid will equal the moles of our weak base. So we are going to consider the missing point corresponding to 30 mL of tyrant. And we're going to do the work for that. Now, so we're considering the addition of 30 mL of hydroponic acid, 0.1 25 moler hydroponic acid hydrant to our three methyl Perine solution. So, doing the work for that below, we're going to need to first begin with finding our initial moles of our base through methyl Perine. So to find that we're going to use its volume and solution given from our prompt as 20 mL, which we will then multiply by its molar or molar concentration given as 0.1 50 molar, which we will interpret as 0.1 50 moles per liter, which are equivalent to molar. Then we are going to recognize that we have inconsistent units of volume, we have liters and millis. So we're going to next multiply by a conversion factor to go from 10 to the third power milliliters in the numerator or rather 10 to the third power millis in the denominator. And that is equivalent to one liter in the numerator. And it's important to set it up that way so that we can now cancel out our units of leaders as well as millis. And that leaves us with moles, which is what we want as our final unit. And we would find that our initial moles uh three methyl period is equal to 0.0030 moles of weak base initially before titration. Next, we need our initial moles of our second reagent, which is our acid hydroponic acid, our tight turn. And so we're given its volume, which we are considering at 30 mL from the prompt, which we will multiply by its molar concentration given as 0.1 25 molar, which we will interpret as 0.1 25 moles per liter. And then again, to make our units of volume consistent, we're going to incorporate the multiplication of our conversion factor where as we stated before 10th of the third power milliliters is equivalent to one leader in the numerator. So again, canceling out mill leaders as well as leaders. We are left with moles and our initial moles of strong I said will equal 0.00375 moles initially of strong acid. So now that we have our initial moles of our reactants, we're going to set up an IC F table to show the final conditions of our reaction once titration occurs. So writing out our reaction again, we have three methyl Perine reacting with hydroponic acid to form three methyl Pyridium ion. And we're only going to include our three methyl perineum ion because as we stated before, our bromide anion is a spectator. So we're not going to consider that in our IC F table. So we can plug in our initial conditions where the eye is. So for our initial conditions, we just calculated that we have initially 0.0030 moles of three methyl pine and 0.00375 moles of strong acid hydroponic acid because our product has not formed yet. Initially, we will have zero of that initially. Next, we're going to recognize that from our calculations of initial moles, we want to determine the limiting reagent. And that is going to be the reagent that has a lower number of moles. So because we have only 0.0030 moles of three methyl Peridin, it is going to be the limiting reagent. And so that is going to be our change for C in our IC F table. So we're going to subtract by that amount from our reagents. So minus 0.0030 moles for both of our reagents and we would have plus that amount for our product. So now for our final conditions, once we apply the subtraction, our final conditions will result in zero moles of our weak base for our strong acid. And after the titration, we will have 0.0075 moles remaining. And then for our base, of course, we have what we form which is 0.0030 moles. Now, next, we want to recall an important fact. We stated that three methyl Perine is a weak base and hydroponic acid is a strong acid. So let's recall that in the titration of a weak base and strong acid after the equivalence point, only the strong acid influences Ph. So although we have our three mesyl Perini C, which is the conjugate acid of our weak base, the strong acid hydroponic acid is going to have the strongest influence on Ph after the equivalence point. So we're going to need to take our next step by finding what the equivalence point is to make sure we've passed that or not. So in order to find the equivalence point, we're going to begin by taking our volume of base, which is listed as 30 mL from the prompt again, multiplied by its molar concentration 0.1 25 and sorry correction, our volume of base was given from our prompt as 20 mL. So let's redo this, we have 20 mL of base multiplied by its smaller concentration of point 1 50 moles per liter. And again, for consistent units of volume, we will incorporate our volume conversion factor where 10th of the third power millis and the denominator is equivalent to one liter. This is set equal to our volume of acid which we will need to solve for at the equivalence point. So volume of hydro brom acid which is then and just to make more room. So our volume of acid that we are solving for is multiplied by its molar concentration 0.1 25 moles per liter, which is then also multiplied by. And actually, we would end the calculation on the right hand side of our equation there so that we can isolate for volume. So we're going to need to on the left hand side of our equation cancel out millis as well as leaders. And we're going to divide both sides of our equation by 0.1 25 moles per liter. And in doing so that term will cancel out on the right. And we would also be able to cancel out the unit moles on the left hand side of our equation, leaving us with leaders in our denominator there. And so simplifying in our next line, we would find that the volume of hydro Bromma acid and will use the color black to keep things clear. So the volume of hydroponic acid is equal to 0.1 024 liters. And then to convert this back to millis, we're going to multiply where we have 10 to the negative third liters in the denominator equivalent to one milli in the numerator canceling out liters. We're left with millis as our next unit of volume. And we would find that we have 24 millis of hydroponic acid at the equivalence point. And so going to our chart, we observe that based on the volume we are calculating for 30 mL, we are certainly past the equivalence point where our equivalence 0.24 millilitres of tyrant occurs around here at a ph between three and four. So it maybe 3.5 we can say is our equivalence point. So let's label that equivalence point in even though it's not really required just so we know. So we are certainly past the equivalence point since we did a calculation for our 30 mL of volume of tiran. Since we know that after the titration with our solution of weak base that we have zero point 0075 moles of hydroponic acid left from our IC F table, we're going to need to convert this to a concentration. So our concentration of hydroponic acid will equal the moles and the numerator that remain 0.0075 moles of HBR which is then divided by the total volume of our solution. Since we're called that molar concentration is in terms of moles per liter. So our total volume will include the 30 mL of tyrant HBR which is then added to the 20 mL of three methyl purine weak base, we'll say, and again, we need consistent units of volume to be in liters. So we're going to multiply to go from 10 to the third millis and the denominator equivalent to one liter in the numerator and canceling out milit, we're left with liters moles per liter rather as our final units. And just to make a quick correction, we are missing 10 from our moles remaining of hydroponic acid. So we should have 0.0075 moles of HBR in both our IC F table for the final condition of HBR, as well as in our numerator for calculating its final concentration. And that will give us our concentration for HBR being zero point 015 molar. And so thus, we can now calculate our Ph since recall that our strong acid is going to dissociate. So we would have protons in solution. And thus, we can recall that PH is calculated by taking the negative log of our concentration of H plus or protons. So we would have the negative log of our concentration, 0.015 molar, which will give us a P equal to the value of 1.82 at 30 mL of volume. So going back to our titration curve, we can plot this ph point where at around 30 mL of volume, we have a ph of 1.82. So we'll draw our next point at this location between two and one and 30 millimeters of volume. And now connecting all of our ph points, we will now have our complete titration curve as our final answer to complete this example. So I hope this made sense and let us know if you have any questions.

A 25.0-mL sample of 0.125 M pyridine is titrated with 0.100 M HCl. Calculate the pH at each volume of added acid: 0 mL, 10 mL, 20 mL, equivalence point, one-half equivalence point, 40 mL, 50 mL. Sketch the titration curve.

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Key Concepts

Titration

pH and pKa

Titration Curve