C6H5NH2 + H2O C6H5NH3+ + OH-
1. Aniline, a weak base, reacts with water according to the rxn above.
a. A sample of aniline is dissolved in water to produce 25.0 ml of a .10 M soln. The pH of soln is 8.82. What is Kb for this rxn?
b. The soln in part b, is titrated with .10 M HCl. Calculate pH of the soln when 5 mL of acid is added.
c. Calculate pH at equivalence point of the titration in part c.
d. The pKA values for several indicators are given below. Which of the indicators is most suitable for this titration. Justify your answer.
Indicator pKa
Erythrosine 3
Litmus 7
thymolphtalein 10
a. Kb = 1.8 x 10^-5
b. pH = 7.82
c. pH = 4.82
d. Thymolphtalein is most suitable for this titration because its pKa value is closest to the equivalence point of the titration (4.82).
To solve this problem step-by-step, let's go through each part one by one:
a. To find Kb, we can start by using the pH of the solution to calculate the concentration of OH- ions:
pOH = 14 - pH
pOH = 14 - 8.82
pOH = 5.18
Now, convert pOH to OH- concentration:
OH- concentration = 10^(-pOH)
OH- concentration = 10^(-5.18)
Since the reaction is in a 1:1 ratio, the concentration of OH- ions is also the concentration of C6H5NH3+ ions.
C6H5NH3+ concentration = 10^(-5.18)
Now, let's calculate the concentration of aniline (C6H5NH2) remaining in the solution after it reacts:
C6H5NH2 concentration = initial concentration - C6H5NH3+ concentration
C6H5NH2 concentration = 0.10 M - 10^(-5.18) M
Now, use the equilibrium expression for the reaction and plug in the values we have:
Kb = [C6H5NH3+][OH-] / [C6H5NH2]
Kb = (10^(-5.18) M)(10^(-5.18) M) / (0.10 M - 10^(-5.18) M)
b. To calculate the pH of the solution after 5 mL of HCl is added, we need to determine the number of moles of HCl added:
Number of moles of HCl = concentration of HCl * volume of HCl
Number of moles of HCl = 0.10 M * (5 mL / 1000) L
Since HCl is a strong acid, it completely dissociates, so the number of moles of OH- ions added is the same as the number of moles of HCl:
Number of moles of OH- ions added = Number of moles of HCl
Since we know the initial concentration of OH- ions, we can calculate the new concentration after the addition:
New OH- concentration = OH- concentration + (Number of moles of OH- ions added / new volume of solution)
Now, convert the OH- concentration to pOH and then to pH to find the final pH of the solution.
c. At the equivalence point of the titration, the moles of HCl added is equal to the moles of C6H5NH2 originally present in the solution. We can calculate the number of moles of C6H5NH2 using its initial concentration and the volume of solution:
Number of moles of C6H5NH2 = initial concentration of C6H5NH2 * volume of solution
Now, divide the moles of C6H5NH2 by the new volume of the solution (which includes the volume of HCl added) to find the concentration of C6H5NH2 at the equivalence point.
Then, use the equilibrium expression to find the concentration of OH- ions at the equivalence point and convert it to pH.
d. To determine the suitable indicator for this titration, we need to look at the pH range over which each indicator changes color. The pH range should include the pH at the equivalence point.
Erythrosine changes color around pH 3, which is too low for this titration.
Litmus changes color around pH 7, which is also not suitable.
Thymolphtalein changes color around pH 10, which is closer to the pH expected at the equivalence point.
Therefore, thymolphtalein is the most suitable indicator for this titration as its pH range includes the pH at the equivalence point of this reaction.
To solve this problem, we need to understand the concept of acid-base reactions and how to calculate the pH and dissociation constant (Kb) of a weak base.
1. Calculation of Kb:
We can use the pH of the solution to calculate Kb. The pH value can be related to the concentration of hydroxide ions (OH-) in the solution. In this case, we know that at equilibrium, the concentration of OH- is equal to the concentration of aniline that has undergone partial dissociation.
a. The given pH of the solution is 8.82. To find the concentration of OH-, we can use the equation:
pOH = 14 - pH
pOH = 14 - 8.82 = 5.18
We know that pOH is equal to the negative logarithm of the hydroxide ion concentration (OH-). Taking the antilog of pOH, we can solve for the concentration of OH-:
[OH-] = 10^(-pOH) = 10^(-5.18)
Since the concentration of OH- is equal to the concentration of aniline (C6H5NH2) that underwent partial dissociation, we now have the concentration of aniline (0.10 M) and the concentration of OH-. We can substitute these values into the Kb expression:
Kb = [C6H5NH3+][OH-] / [C6H5NH2]
b. To calculate the pH when 5 mL of HCl is added, we need to consider the stoichiometry of the reaction. The reaction between aniline and HCl is a neutralization reaction, resulting in the formation of an acidic salt. The reaction consumes OH- ions in the solution, consequently reducing the pH.
We need to determine how many moles of aniline react with 5 mL of 0.10 M HCl. From the balanced equation:
1 mole of aniline reacts with 1 mole of HCl
Since we have 0.10 M HCl, we can calculate the moles of HCl used:
moles HCl = concentration (M) x volume (L) = 0.10 M x 0.005 L = 0.0005 moles
Since the stoichiometry is 1:1, the moles of aniline consumed will also be 0.0005 moles. We subtract this amount from the initial moles of aniline to find the remaining moles. From there, we can calculate the new concentration of aniline, and finally, the new pH.
c. When the titration reaches the equivalence point, it means that the moles of HCl added are stoichiometrically equivalent to the moles of aniline present initially.
To find the equivalence point, we need to determine how many moles of HCl are required to react with all the aniline. From the balanced equation:
1 mole of aniline reacts with 1 mole of HCl
The moles of aniline initially present are:
moles C6H5NH2 = concentration (M) x volume (L) = 0.10 M x 0.025 L = 0.0025 moles
Therefore, the moles of HCl required for the titration will also be 0.0025 moles. From there, we can calculate the volume of HCl needed to reach the equivalence point. This volume will allow us to calculate the new concentration of OH- and, consequently, the new pH.
d. To determine the most suitable indicator for a titration, we need to consider the pH range over which the color change of each indicator occurs. The indicator should change color around the pH at the equivalence point.
When selecting an indicator, we want its pKa to be close to the expected pH at the equivalence point. In this case, since we have an acidic solution due to the HCl, we would expect the equivalence point to be around pH 7 or lower.
Considering the pKa values given:
Erythrosine (pKa 3): This indicator will change color in a very acidic range and is not suitable for this titration.
Litmus (pKa 7): This could be a suitable indicator as it changes color near pH 7, which is close to the expected pH at the equivalence point.
Thymolphtalein (pKa 10): This indicator is not suitable as it changes color at a much higher pH than expected for the equivalence point.
Therefore, Litmus is the most suitable indicator since it changes color around the pH range expected at the equivalence point (pH 7).