A buffered solution is made by adding 47.4 g C2H5NH3Cl to 1.00 L of a 0.68 M solution of C2H5NH2. Calculate the pH of the final solution. (Assume no volume change. Assume that all solutions are at 25°C.)

Convert 47.4 g C2H4NH3Cl to mols. mols = grams/molar mass = ?. Since that is in 1 L solution, that will be the molarity. This is the acid.

You have 0.68 M solution of C2H4NH2 which is the base.
Substitute into the Henderson-Hasselbalch equation, and solve for pH.

To calculate the pH of the final solution, we need to determine the concentrations of the acidic and basic components of the buffer system.

1. Begin by calculating the number of moles of C2H5NH2 in the 0.68 M solution. To do this, multiply the molarity (0.68 M) by the volume (1.00 L):
moles C2H5NH2 = 0.68 M x 1.00 L = 0.68 moles

2. Convert the mass of C2H5NH3Cl (47.4 g) to moles using its molar mass. The molar mass of C2H5NH3Cl can be calculated as follows:
Molar mass = (2 x molar mass of C) + (5 x molar mass of H) + molar mass of N + 3 x molar mass of H + molar mass of Cl
= (2 x 12.01 g/mol) + (5 x 1.01 g/mol) + 14.01 g/mol + (3 x 1.01 g/mol) + 35.45 g/mol
= 107.46 g/mol

moles C2H5NH3Cl = mass (g) / molar mass (g/mol)
= 47.4 g / 107.46 g/mol
= 0.441 moles

3. Now, consider the reaction that occurs between C2H5NH2 (a weak base) and C2H5NH3Cl (its conjugate acid):
C2H5NH2 + HCl ↔ C2H5NH3Cl
The equilibrium expression for this reaction is:
Ka = [C2H5NH3Cl] / [C2H5NH2]

Since we know the initial concentration of C2H5NH2 is 0.68 M, and we are adding 0.441 moles of C2H5NH3Cl to 1.00 L of solution, we can calculate the final concentration of C2H5NH2:
final concentration of C2H5NH2 = (moles initial C2H5NH2 - moles C2H5NH3Cl) / volume
= (0.68 moles - 0.441 moles) / 1.00 L
= 0.239 M

Now, we need to calculate the concentration of C2H5NH3Cl in the final solution:
final concentration of C2H5NH3Cl = moles C2H5NH3Cl / volume
= 0.441 moles / 1.00 L
= 0.441 M

4. Now that we have the concentrations of C2H5NH2 and C2H5NH3Cl, we can calculate the pH of the final solution. The pH of a buffered solution can be calculated using the Henderson-Hasselbalch equation:
pH = pKa + log([base] / [acid])

The pKa can be determined from the Ka value:
Ka = [H+][C2H5NH2] / [C2H5NH3Cl]
pKa = -log(Ka)

For C2H5NH2 (a weak base), pKa = 14 - pKb
pKb = -log(Kb)

We can assume that Kb = Kw / Ka(H2O), where Kw is the ionization constant for water (1.0 x 10^-14) and Ka(H2O) is the autoionization constant for water (also 1.0 x 10^-14).

Kb = Kw / Ka(H2O) = 1.0 x 10^-14 / 1.0 x 10^-14 = 1.0

pKb = -log(1.0) = 0

Therefore, pKa = 14 - 0 = 14

Plugging the values into the Henderson-Hasselbalch equation:
pH = 14 + log([0.239] / [0.441])

Calculate the natural logarithm of [0.239] / [0.441] using a calculator:
ln([0.239] / [0.441]) = -0.639

pH = 14 + (-0.639)
= 13.361

Therefore, the pH of the final solution is approximately 13.361.