Consider the titration of a 25.0-mL sample of 0.175 M CH 3 NH 2 with 0.150 M HBr. Determine each quantity. c. the pH at 5.0 mL of added acid

All right. Hello everyone. So this question says to consider the titration of a 25.0 mL sample of 0.175 molar methyl amine with 0.150 molar HBR determine each quantity. And for this part, we have the P at 5.0 mL of added acid. Here are four different answer choices, proposing different values for the PH. So for this question, right, we're going to see a reaction between a weak base that's methyl amine and HBR which is a strong acid. So to calculate how many moles of reactant and product are present before and after the reaction, we're going to have to use an IC F table. But before we do that, we do have to find the initial concentrations of both the weak base and the strong acid. So let's go ahead and do that. Starting off with the concentration of methyl amine thats CH three NH two, right. The first thing we're going to do is divide the molar concentration, excuse me, multiply the molar concentration by the volume of the sample. However, before we do that, I do just want to very quickly convert the volume of the sample 2 L because we're given that this quantity in milliliters because recall that molar concentration is expressed as moles per liter. So to keep units consistent, I'm going to convert 25.0 mL. Now recall that this is done by dividing the quantity in milliliters by 1000. Or we can simply take the decimal place and move it three spaces to the left. So if I take my molar concentration, that's 0.175 moles per liter, I'm going to multiply this by the volume that I've since converted into liters, but zero point 0 2 5 L of the sample. And so this gives me 0.004375 moles of methyl amine at the beginning of the reaction. So now we're going to do the same thing for the moles of HBR that have been added in the scenario. So for the concentration of H pr in the beginning, I'm going to take its molder concentration, that's 0.150 moles per liter. And I'm going to multiply this by 5.0 mL divided by 1000. So that's 0.005 L. This ensures that my units of leaders cancel out in both cases. So for the initial concentration of HBR, we get 0.00075 moles. So now these are the values that were going to start off with in our ICF table So the acid base reaction equation is as follows. First, we have methyl amine in aqueous solution reacting with HBO that is also an aqueous solution combining to create their conjugate acid and conjugate base respectively. Right after an acid base reaction between these two compounds, the nitrogen and methyl amine is going to gain an additional proton. And so the conjugate acid is going to look very similar, but nitrogen is going to have one extra proton attached to it and therefore a positive charge. On the other hand, because HBR is an acid, it is going to lose the proton that it has. And after that happens, the conjugate base is going to be simply a bromide ion. So now on the side here, I'm going to add labels to the I of the C and the F ropes, I stands for initial concentration, C stands for change over the course of the reaction and F stands for final concentration. So in this case, right, for the first row of my table that represents the initial concentrations, we're going to use the number of moles of both meth amine and H pr that we calculated previously. So for meth amine, that would be 0.004375 moles. And for H pr that will be 0.00075 moles. Now, at the very beginning of the reaction, there will be no products in solution quite yet. Though the concentration of methyl amine or sorry, the conjugate acid of methyl amine should be zero moles. And one quick note about bromide, which is the conjugate base of HBR because recall that bromide is the conjugate base of a strong acid. This means that it is a neutral ion that will not actually contribute to the PH. So because we're interested in finding the PH, we can disregard the concentration of bromide because it is a neutral ion. So with this in mind, let's go ahead and discuss the change in concentrations over the course of this reaction. Now, H PR is a strong acid, whereas meth amine is a weak base, this means that throughout the course of the reaction, only HBR gets consumed entirely because HBR is the species that can dissociate completely in water. It is a strong acid. So in the change column, excuse me, change row, right, positive and negative values are used to indicate whether or not that species is being created or consumed throughout the process. So in the case of HBR, right, it is going to be consumed completely. So there should be a negative sign for its value in the change column. And so because it's going to be consumed completely, I'm going to subtract the number of moles by itself to give me zero at the very end. Now, because meth amine is also a reactant, it is also going to be consumed or lost throughout the reaction. So here I'm going to subtract the moles of methyl amine by the same amount in the change column. However, the conjugate acid of methyl amine is a product which is going to be formed, right. So instead of adding a negative sign implying that its being subtracted from, I am going to add a positive side and I am going to add 0.00075 moles to the initial concentration of the conjugate acid because once again, it is a product so its concentration should increase. All right. So at this point for the row representing the final concentrations, we're going to simply add or subtract the values in the I and the C columns for each compound. Now because HBR gets consumed completely, there will be zero bowls of HBR. After the reaction for the conjugate acid of methyl amine, there are going to be 0.00075 moles at the very end. And as for methyl amine, we're going to subtract the change in concentration from the initial concentration that we calculated in the beginning. And so that gives us 0.00 3625 moles of meth amine after the reaction. So now let me scroll down because it's noteworthy here that after the acid base reaction that's being described, the only species that will be present are the weak base thats methyl amine and its conjugate acid, right? The strong acid is no longer in solution. And this means that we're going to end up with a buffer solution and recall that the PH of a buffer solution can be calculated using the Henderson Hasselback equation in which the PH is equal to the PK of the compound added to the logarithm of the concentration of base over the concentration of acid. However, there is a caveat in the sense that methyl amine is a base, which means that we know the KB of methyl amine. However, using the KW, we can calculate the K A from the KB. So according to either an online source or the textbook itself, right, the KB of methyl amine is 4.4 multiplied by 10 to the power of negative four. And so KW is equal to the K A multiplied by the KB. So recall that the KW at 25 °C is 1.0 multiplied by 10 to the power of negative 14. So K A is equal to KW divided by KB. So that's 1.0 multiplied by 10 to the power of negative 14, divided by KB which was 4.4 multiplied by 10 to the negative fourth. And this results in A K A value of 2.2727 multiplied by 10 to the power of negative 11. So now this is the K A value that we can plug into our Henderson Hasselbach equation. However, because the PK is required as opposed to the K A recall that the PK is equal to the negative logarithm of the K A value. So in the Henderson Hasselbach equation, the PH is going to be equal to the negative logarithm of the K A. That's 2.2727 mortified by 10 to the power of negative 11 added to the logarithm of the concentration of base divided by the concentration of acid. So that's the logarithm of 0.003625 moles because in this case, concentration is expressed in moles. Now, the reason for this is because the acid and the base will eventually have the same final volume. So we can go ahead and use moles in this case. And so the moles of acid is 0.00075. So when I evaluate this expression all the way through my PH is equal to 11.33. Now I'm rounding off to two decimal places because the smallest number of significant figures from the given values was two and there you have it. So our ph is 11.33 which corresponds to option A in the multiple choice. So if you watch this video all the way through, I thank you so very much for watching and I hope you found this helpful.

Ch.18 - Aqueous Ionic Equilibrium

All textbooks

Tro 6th Edition

Ch.18 - Aqueous Ionic Equilibrium

Problem 79c

Chapter 18, Problem 79c

Consider the titration of a 25.0-mL sample of 0.175 M CH₃NH₂ with 0.150 M HBr. Determine each quantity. c. the pH at 5.0 mL of added acid

Verified step by step guidance

Identify the initial moles of CH_3NH_2 in the solution using the formula: \( \text{moles} = M \times V \).

Calculate the moles of HBr added using the formula: \( \text{moles} = M \times V \).

Determine the moles of CH_3NH_2 remaining after reaction with HBr by subtracting the moles of HBr from the initial moles of CH_3NH_2.

Calculate the concentration of CH_3NH_2 and CH_3NH_3^+ in the solution after the reaction, considering the total volume of the solution.

Use the Henderson-Hasselbalch equation to find the pH: \( \text{pH} = pK_a + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right) \), where \( pK_a \) is derived from \( K_b \) of CH_3NH_2.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Titration

Titration is a quantitative analytical technique used to determine the concentration of a solute in a solution. It involves the gradual addition of a titrant (a solution of known concentration) to a sample until a reaction reaches its endpoint, often indicated by a color change. In this case, the titration involves a weak base (methylamine) being neutralized by a strong acid (HBr).

pH Calculation

pH is a measure of the acidity or basicity of a solution, calculated as the negative logarithm of the hydrogen ion concentration. In titrations, the pH can change significantly as the titrant is added, especially near the equivalence point. For weak bases like methylamine, the pH at various points can be calculated using the Henderson-Hasselbalch equation or by considering the dissociation of the weak base and the resulting conjugate acid.

Buffer Solutions

Buffer solutions are mixtures that resist changes in pH upon the addition of small amounts of acid or base. In the context of this titration, when a weak base is partially neutralized by a strong acid, a buffer system is formed, which can help maintain a relatively stable pH in the solution. Understanding how buffers work is crucial for predicting the pH at specific points during the titration.