A buffer solution is prepared by dissolving 0.400 mol of CH3COOH and 0.200 mol of CH3COONa in 1.00 L of water. 1.00 mL of 10.0 M HCl is added to a 100 mL portion of this solution. What is the final pH of the resulting solution?

Question

A buffer solution is prepared by dissolving 0.400 mol of CH3COOH and 0.200 mol of CH3COONa in 1.00 L of water. 1.00 mL of 10.0 M HCl is added to a 100 mL portion of this solution. What is the final pH of the resulting solution?

Ka of CH3COOH = 1.8 x 10¯5

Answer 1

See the post just above yours.

Answer 2

To find the final pH of the resulting solution, we need to consider the effects of the added HCl on the buffer solution.

First, let's calculate the initial concentrations of CH3COOH and CH3COONa in the buffer solution.

For CH3COOH:
Number of moles = 0.400 mol
Volume of solution = 1.00 L
Concentration (C) = moles/volume = 0.400 mol / 1.00 L = 0.400 M

For CH3COONa:
Number of moles = 0.200 mol
Volume of solution = 1.00 L
Concentration (C) = moles/volume = 0.200 mol / 1.00 L = 0.200 M

Now, let's calculate the concentration of HCl that was added to the solution.

For HCl:
Volume added = 1.00 mL = 0.001 L
Concentration (C) = 10.0 M

The moles of HCl added = volume x concentration = 0.001 L x 10.0 M = 0.010 mol

Next, we need to calculate the new concentrations of CH3COOH and CH3COONa after the addition of HCl.

For CH3COOH:
The original moles of CH3COOH = 0.400 mol
The moles of CH3COOH that reacted with the HCl = 0.010 mol (stoichiometric ratio is 1:1)
The remaining moles of CH3COOH = original moles - reacted moles = 0.400 mol - 0.010 mol = 0.390 mol

The new concentration of CH3COOH = remaining moles / volume = 0.390 mol / 1.00 L = 0.390 M

For CH3COONa:
The original moles of CH3COONa = 0.200 mol
The moles of CH3COONa do not change upon reaction with HCl, as it is a neutral salt. Therefore, the concentration remains the same at 0.200 M.

Now, let's determine the concentration of H3O+ ions in the solution based on the Ka value of CH3COOH.

The equation for the dissociation of CH3COOH is:
CH3COOH + H2O ⇌ CH3COO- + H3O+

Since the number of moles of CH3COOH and CH3COO- are equal in the buffer solution (0.390 mol), the concentration of CH3COO- is also 0.390 M.

Using the equation for Ka:
Ka = [CH3COO-][H3O+] / [CH3COOH]

Substituting the values:
1.8 x 10^-5 = (0.390 M)([H3O+]) / 0.390 M

Simplifying the equation, we get:
[H3O+] = 1.8 x 10^-5 M

Finally, to find the pH of the solution, we use the equation:
pH = -log[H3O+]

pH = -log(1.8 x 10^-5)
pH ≈ 4.74

Therefore, the final pH of the resulting solution is approximately 4.74.