So this is a three part question. I solved the first part correctly but I need help with the second and third.
Q: The dissociation equilibrium constants for the protonated form of of alanine (a diprotic amino acid, H2X+)are Ka1 = 4.6 x 10^-3 and Ka2 = 2.0 x 10^-10. Assume that you start with 50.0 mL of 0.050 M solution of alanine.
a) What is the pH of this initial solution? My answer: pH = 1.88
b) What is the concentration of HX after 25.0 mL of 0.100 NaOH has been added to the solution?
c) What is the pH of the solution after 25.0 mL of 0.100 NaOH has been added to the solution?
I need more help understanding how to do the actual problem on a) and b) than the actual answer. If someone can set me on the right path I can do all the math to solve it. Thanks.
c. You have 0.05M x 50 mL = 2.5 mmols of the amino acid. When you have titrated with 25 mL of 0.1 M NaOH, you have neutralized all of the COOH part but none of the amine part and you have the zwitter ion. That is the first equivalence point (or the isoelectric point and pH = (pka1 + pka2)/2.
b. You start with 2.5 mmols alanine. When all has been converted to the zwitter ion by the addition of the 25 mL of 0.1 M NaOH, the concn of that material then is 25 mmols/75 mL = ?M.
To solve part (a) of the problem, you need to find the pH of the initial solution. The pH of a solution can be calculated using the equation:
pH = -log[H+]
where [H+] is the concentration of hydrogen ions in the solution. In this case, the protonated form of alanine, H2X+, is dissociated into two hydrogen ions, H+.
Given that the initial solution is a 0.050 M solution of alanine, the concentration of H2X+ is also 0.050 M. However, since it is diprotic, it dissociates in two steps.
The first dissociation reaction is:
H2X+ ⇌ HX- + H+
The equilibrium constant for this reaction is Ka1 = [HX-][H+]/[H2X+].
Using the information given, Ka1 = 4.6 x 10^-3.
We can assume that at equilibrium, the concentration of [HX-] and [H+] will be the same, which we'll call x M.
So, at equilibrium, [HX-] = x M and [H+] = x M.
[H2X+] = initial concentration - [HX-] = 0.05 M - x M.
Since x is very small compared to 0.05 M, we can approximate [H2X+] as 0.05 M.
Now, we can substitute these concentrations into the equilibrium constant expression:
Ka1 = [HX-][H+]/[H2X+]
4.6 x 10^-3 = (x)(x)/(0.05)
Next, solve for x using quadratic equation or by assuming x is small relative to 0.05 M.
By approximating x as small relative to 0.05 M, we can ignore x in the denominator, which simplifies the equation:
4.6 x 10^-3 ≈ x^2/0.05
Rearranging the equation, we have:
x^2 ≈ (4.6 x 10^-3)(0.05)
x ≈ sqrt[(4.6 x 10^-3)(0.05)]
Solve for x to get the concentration of [H+] and then calculate the pH using the pH formula.
For part (b) of the problem, after 25.0 mL of 0.100 NaOH is added to the solution, it will react with the available HX- in a 1:1 stoichiometric ratio:
NaOH + HX- ⇌ NaX + H2O
Since the initial concentration of HX- is given by the equilibrium concentration at the end of part (a), you can calculate the remaining concentration of HX- after 25.0 mL of NaOH is added.
To calculate the concentration of HX-, use the initial concentration of HX- minus the concentration of NaOH that reacted (converted to moles):
[HX-] = Initial concentration of HX- - Concentration of NaOH reacted
Finally, for part (c), after finding the concentration of HX-, you can use the concentration of HX- to calculate the concentration of OH- due to the NaOH that reacted with HX- in part (b). From there, you can find the pOH using the equation:
pOH = -log[OH-]
Then, calculate the pH using the equation:
pH = 14 - pOH
Sure! I'll guide you through solving parts b) and c) of the question.
Part b) asks for the concentration of HX after 25.0 mL of 0.100 M NaOH has been added to the solution. To solve this, we need to consider the reaction that occurs between HX and NaOH. The balanced equation for the reaction is:
HX + NaOH -> NaX + H2O
In this reaction, HX reacts with NaOH to form NaX (a salt) and water.
Now, let's think about the moles of NaOH that have been added. We know that the volume of the NaOH solution that has been added is 25.0 mL, and the concentration is 0.100 M. Using the formula C1V1 = C2V2, where C1 and V1 are the initial concentration and volume, and C2 and V2 are the final concentration and volume, we can calculate the moles of NaOH added:
moles of NaOH = (0.100 M) * (25.0 mL / 1000) = 0.0025 moles
Since HX is a diprotic amino acid, we can assume that equal amounts of NaOH react with HX. Therefore, 0.0025 moles of HX will react.
To find the final concentration of HX, we need to consider the initial concentration of HX before any NaOH has been added. You mentioned that the initial solution contains 50.0 mL of 0.050 M alanine. Since alanine is diprotic, we have to consider that half of the alanine molecules are protonated, i.e., H2X+.
Therefore, the initial moles of H2X+ can be calculated as follows:
moles of H2X+ = (0.050 M) * (50.0 mL / 1000) = 0.0025 moles
Subtracting the moles of H2X+ that reacted with NaOH from the initial moles, we get:
moles of HX remaining = 0.0025 - 0.0025 = 0 moles
Since all the HX molecules have reacted, the concentration of HX is 0 M.
Moving on to part c), we need to find the pH of the solution after 25.0 mL of 0.100 M NaOH has been added. Since all the HX has reacted, we only need to consider the NaOH that has been added to the solution. The amount of NaOH added to the solution is 0.0025 moles, and the total volume of the solution after the addition is 75.0 mL (initial volume + NaOH volume).
To find the concentration of OH- in the solution, we divide the moles of NaOH by the total volume:
concentration of OH- = 0.0025 moles / (75.0 mL / 1000) = 0.033 M
Using the equation for the ionization of water, KW = [H+][OH-], we can find the concentration of H+:
[H+] = KW / [OH-] = (1.0 x 10^-14) / (0.033 M) ≈ 3.03 x 10^-13 M
To find the pH, we take the negative logarithm of the concentration of H+:
pH = -log[H+] ≈ -log(3.03 x 10^-13) ≈ 12.52
Therefore, the pH of the solution after adding 25.0 mL of 0.100 M NaOH is approximately 12.52.
I hope this explanation helps you understand how to approach the problem. If you have any further questions, feel free to ask!