what is the pH of a solution that is 0.40 M in sodium acetate and 0.60 M on acetic acid? (Ka for acetic acid is 1.85x10^-5.)
To find the pH of the solution, we can use the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation relates the pH of a solution to the concentration of an acid and its conjugate base.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
Where:
pH = the pH of the solution
pKa = the negative logarithm of the acid dissociation constant of the acid (Ka)
[A-] = the concentration of the conjugate base
[HA] = the concentration of the acid
In this case, acetic acid (CH3COOH) is the acid and sodium acetate (CH3COONa) is the conjugate base. The Ka for acetic acid is given as 1.85x10^-5.
First, we need to calculate the concentrations [A-] and [HA] using the given molarities (M) of the solutions.
[A-] = 0.40 M (concentration of sodium acetate)
[HA] = 0.60 M (concentration of acetic acid)
Now, substitute these values into the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
= -log(1.85x10^-5) + log(0.40/0.60)
Calculating the logarithms and division:
pH = -log(1.85x10^-5) + log(0.67)
= -(-4.73) + (-0.18)
Finally, calculating the addition:
pH = 4.73 - 0.18
Therefore, the pH of the solution is approximately 4.55.