Two 20.0-mL samples, one 0.200 M KOH and the other 0.200 M CH₃NH₂, are titrated with 0.100 M HI. d. Sketch each titration curve.

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Welcome back everyone. 2 25 mL basic samples, 1.2 50 molar lithium hydroxide and the other 500.2 50 molar hyla were titrate with 500.1 25 molar hydro brom acid complete the titration curve for 250.2 50 molar lithium hydroxide. If a missing point corresponds to 62.5 mL of 0. 25 molar hydro brom acid. Given the titration curve for 0.2 50 molar hyaline compare the equivalence points below. We are given two titration curves. Our first being for lithium hydroxide which is incomplete and only contains six points where we are given a marked equivalence point at a ph of seven at mL of tyrant. For our second titration curve, we have a complete curve for the titration of ethyl. And we observe that we have a p of about six at milli of tirin. So we're going to need to take our first step by finding that missing point for our titration of lithium hydroxide. So we're going to begin by writing out our reaction between lithium hydroxide and hydro Brom acid. Recall that lithium hydroxide is on our memorized list of strong bases in our textbook, as well as hydroponic acid is in our memorized list of strong acids from our textbooks. So we have a strong base reacting with a strong acid which will result in a neutralization reaction and recall that the products will be a salt and then water our salt and for water, we would just have H2O. So our salt is going to form from the combination of our cat ion from our base and our anion from our acid. So that would give us lithium bromide. And based on our soluble rules for bromides, this is a soluble salt. So it's aqueous. Now, we also should take note that we have a 1 to 1 molar ratio between our strong base and strong acid. And therefore recall that this tells us that at the equivalence point for this neutralization reaction, our moles of acid will equal our moles of base. We should also recall that for this neutralization reaction at the equivalence point, the P H should equal seven, which we do observe our given equivalence point which is plotted at a ph of seven and 50 mL of tyrine. So now with this understood, we're going to find that missing ph at the volume of 62.5 mL of acid 0.1 25 molar acid, which is type trading hour 0.2 50 molar lithium hydroxide solution. So we're going to need to begin by finding our initial amounts of our reactants. So our initial moles of lithium hydroxide. We can start with, we would find this by taking its volume 25 millis from the prompt and multiplying by its molar concentration 250.2 50 which we will interpret as moles per liter since that is equivalent to molar. And notice that we have inconsistent units of volume. So we will now multiply by our conversion factor to go from 10 to the third power milliliters in the denominator equivalent to one liter in the numerator. So we can now cancel out our units of milliliters as well as liters leaving us with moles. And we would find that our initial moles of lithium hydroxide is equal to 0.625 moles of strong base. Next, we're going to need to calculate our initial moles of acid hydro brom acid. We're going to begin with its volume given us 62.5 mL, which we will then multiply by its molar concentration, which we will interpret as 0.1 25 moles per liter. And then of course, to make our units of volume consistent, we will multiply to go from 10 to the third power millilitres in the denominator equivalent to one liter in the numerator. And again, canceling out our units of liters and millis we're left with moles. And we would find that our initial moles of acid is equal to 0.78125 moles of strong acid Now, with our initial moles of our reactants, we're going to set up an IC F table where we would write our reaction above. So we can just paste in our reaction. And we're only going to notice that we did not include water because that is not considered in an IC F table. We would only be considering our aqueous reents. So for our initial conditions, we just determined that our initial moles of base is 0.625 moles. Our initial moles of acid we just determined was 0.78125 moles. And our initial moles of product, lithium bromide salt is zero since it has not formed yet. Initially. Now, for the change, we're going to need to determine our limiting reactant based on the smaller number of moles, our lower number of moles is associated with our base as 0.625 moles. And so we're going to have the change be minus that amount for our reactants and plus 0.625 moles for the product. So now for our final conditions, we will have no base left after the titration. So we have zero moles left of our tyrant, sorry of lithium hydroxide. For our tiran hydroponic acid, we would have 0.15625 moles remaining. And then of course, for our salt, we would gain or form 0. moles of our salts. So now what really matters is that our salt is going to just associate into ions, lithium and bromide. So we would just be left mainly with our strong acid influencing P H. So now we're going to need to interpret the remaining moles of acid in terms of a concentration for our acid. So recall that we can calculate the concentration by taking our moles of our substance. 0.1. So 0.15625 moles of hydroponic acid that we have remaining. And this is going to be divided by the total volume of our solution since we need a concentration. So our volume of of base 25 mL, lithium hydroxide is added to our volume of acid. 62.5 mL of hydroponic acid both given from the prompt and then we need to have units of liters in our denominators. So we're going to multiply by the conversion factor to go from one mil liter in the denominator equivalent to 10 to the negative third liters in the numerator. So canceling out our units of millis, we're left with moles per liter as our units for molar or concentration. And we would find that our concentration of hydro brom acid is equal to 0.178571 molar of H B R. Now with our molar concentration remaining in solution, we're going to be able to recognize that it will impact Ph because we can calculate P H by taking the negative log of our concentration of protons H plus. And because our strong acid will dissociate into a proton as well as our bromide anion, which is a spectator. We can just plug in the negative log of our concentration of H B R which we determine to be 0.178571 molar, which would be equal to our concentration of H plus. Since again, it's a or since we have one mole of our acid. So now we're just going to find that our P H and our calculators is equal to a value of 1.748, which we can round to about 1.75 as the ph at 62.5 mL of tyrann. And now we're just going to record that onto our titration curve for lithium hydroxide. So at about 62.5 mL, which we'll say is around here, we have a volume or sorry, a P H of about one point 75. So we'll say that the dot will be there. And now connecting all of our P H points, we will form our titration curve for lithium hydroxide, which will be our first and complete answer for this example. Now we just need to compare the equivalence points. So we note that our equivalence point for lithium hydroxide, which is a strong base reacting with its tirin. A strong acid should be a P H of seven at 50 milit, which makes sense again, because this was a neutralization. Whereas our P H for the equivalence point of ethylene was a little under six, but just about six at 15 millis of tyrant. However, we should recognize that ethylone is a weak base. So it's a weak base. And so we have etm a weak base C H three C H two N H two reacting with hydroponic acid, a strong acid. And recall that our products here would be from the proton being donated from our acid to our weak base. Since it will not dissociate, we would have our conjugate acid of Elaine, which would be Hollomon ca ion C H three C H two N H three plus as well as our bromide anion. Now, although you would think that our conjugate acid of our weak base Eloi Calion would influence Ph, we're going to recall that after the equivalence point in a titration with or we should say titration of a weak base with a strong acid, only the strong acid will have the greatest impact on P H. So we can observe that the equivalence point being six for the titration of Elaine continues to lower below seven until we reach the end of our titration for et Lamine at a volume of 100 mL of tyrant with a P H of about one or just under one or actually it is exactly at one. So that makes sense since again, we have ethane, a weak base being reacted with a strong acid. And so this strong acid is the only main component that is influencing P H. So even though there is conjugate acid present, it is the strong acid hydroponic acid that is lowering the P H after the equivalence point. So that is how we would compare our P H points between the two titration curves for our second final answer. So overall, I hope that this made sense and let us know if you have any questions.

Two 20.0-mL samples, one 0.200 M KOH and the other 0.200 M CH₃NH₂, are titrated with 0.100 M HI. d. Sketch each titration curve.

Verified Solution

Key Concepts

Titration

Acid-Base Reactions

Titration Curves

Two 20.0-mL samples, one 0.200 M KOH and the other 0.200 M CH3NH2, are titrated with 0.100 M HI. d. Sketch each titration curve.

Verified Solution

Key Concepts

Titration

Acid-Base Reactions

Titration Curves

Two 20.0-mL samples, one 0.200 M KOH and the other 0.200 M CH₃NH₂, are titrated with 0.100 M HI. d. Sketch each titration curve.