A solution that is prepared by mixing 20.0 mL of 0.32 M HC2H3O2 with 20.0 mL of 0.36 M NaC2H3O2 would have a pH of ___. For HC2H3O2, Ka = 1.8 x 10-5.

Use the Henderson-Hasselbalch equation.

the salt is the base.

Does 4.7 sound correct?

Close but I wouldn't round it off to 4.7.

To find the pH of the solution, we need to consider the acidity of the components present in the solution.

First, let's analyze the components of the solution:

HC2H3O2 (acetic acid) and NaC2H3O2 (sodium acetate) form a buffer solution when mixed together. A buffer solution consists of a weak acid and its conjugate base or a weak base and its conjugate acid.

The dissociation of acetic acid, HC2H3O2, can be represented by the equation:

HC2H3O2 β‡Œ H+ + C2H3O2-

The acid dissociation constant (Ka) for acetic acid is given as 1.8 x 10^-5.

Now, let's calculate the amount of the species in solution after mixing the two solutions:

For HC2H3O2:
Concentration (C1) = 0.32 M
Volume (V1) = 20.0 mL = 0.020 L

moles of HC2H3O2 = C1 * V1
= 0.32 M * 0.020 L
= 0.0064 mol

For NaC2H3O2:
Concentration (C2) = 0.36 M
Volume (V2) = 20.0 mL = 0.020 L

moles of C2H3O2- = C2 * V2
= 0.36 M * 0.020 L
= 0.0072 mol

Since the stoichiometry between HC2H3O2 and C2H3O2- is 1:1, the concentrations of both species will be the same in the final mixed solution.

Now let's calculate the concentrations of H+ and C2H3O2- ions in the solution using the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

Given:
pKa = -log(Ka) = -log(1.8 x 10^-5) = 4.74
[A-] = concentration of C2H3O2- = 0.0072 mol / 0.040 L = 0.18 M
[HA] = concentration of HC2H3O2 = 0.0064 mol / 0.040 L = 0.16 M

Using these values in the Henderson-Hasselbalch equation:

pH = 4.74 + log(0.18/0.16)
= 4.74 + log(1.125)
= 4.74 + 0.05
= 4.79

Therefore, the pH of the solution after mixing 20.0 mL of 0.32 M HC2H3O2 with 20.0 mL of 0.36 M NaC2H3O2 would be approximately 4.79.