Calculate the pH of each of the following buffered solutions:
a. 0.10M C2H5NH2/0.25M C2H5NH3Cl
b. 0.25M C2H5NH2/0.50M C2H5NH3Cl
c. 0.50M C2H5NH2/0.50M C2H5NH3Cl
Mlszwski???
Use the Henderson-Hasselalch equation.
To calculate the pH of a buffered solution, you need to consider the equilibrium between the weak acid/base pair in the buffer solution. The Henderson-Hasselbalch equation can be used for this purpose:
pH = pKa + log ([A-]/[HA]),
where pKa is the acid dissociation constant of the weak acid (A-) and [A-] and [HA] are the concentrations of the conjugate base and acid, respectively.
For the given buffered solutions:
a. 0.10M C2H5NH2/0.25M C2H5NH3Cl:
The weak acid here is C2H5NH2, which is an amine. The conjugate base is C2H5NH3Cl.
The pKa of C2H5NH2 is 10.65.
Substituting the given values into the Henderson-Hasselbalch equation:
pH = 10.65 + log(0.25/0.10)
b. 0.25M C2H5NH2/0.50M C2H5NH3Cl:
Using the same pKa (10.65) for C2H5NH2:
pH = 10.65 + log(0.50/0.25)
c. 0.50M C2H5NH2/0.50M C2H5NH3Cl:
Using the same pKa (10.65) for C2H5NH2:
pH = 10.65 + log(0.50/0.50)
To calculate the pH for each solution, perform the logarithmic calculation using a calculator or computer software.
To calculate the pH of a buffered solution, you need to consider the acid and its conjugate base in the solution. In this case, the acid is C2H5NH3+ (ethanamine) and the conjugate base is C2H5NH2 (ethylamine). The Henderson-Hasselbalch equation is commonly used to calculate the pH of a buffered solution:
pH = pKa + log([A-]/[HA])
where pKa is the negative logarithm of the acid dissociation constant, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.
a. 0.10M C2H5NH2/0.25M C2H5NH3Cl:
First, we need to determine the pKa value for C2H5NH3+ (ethylamine). The pKa of ethylamine is approximately 10.75.
Using the Henderson-Hasselbalch equation:
pH = 10.75 + log([C2H5NH2]/[C2H5NH3Cl])
pH = 10.75 + log(0.10/0.25)
pH = 10.75 + log(0.40)
pH ≈ 10.75 - 0.40
pH ≈ 10.35
b. 0.25M C2H5NH2/0.50M C2H5NH3Cl:
Using the same pKa value of 10.75:
pH = 10.75 + log([C2H5NH2]/[C2H5NH3Cl])
pH = 10.75 + log(0.25/0.50)
pH = 10.75 + log(0.50)
pH ≈ 10.75 - 0.30
pH ≈ 10.45
c. 0.50M C2H5NH2/0.50M C2H5NH3Cl:
Again, using the pKa value of 10.75:
pH = 10.75 + log([C2H5NH2]/[C2H5NH3Cl])
pH = 10.75 + log(0.50/0.50)
pH = 10.75 + log(1)
pH ≈ 10.75
So, the pH values of the buffered solutions are approximately:
a. pH ≈ 10.35
b. pH ≈ 10.45
c. pH ≈ 10.75