A 0.1 M solution of acetic acid is titrated with 0.05M solution of NaOH. What is the pH when 60% of the acid has been neutralized? The equilibrium constant (Ka) for acetic acid is 1.8x10^-5
You can do this one of two ways, both using the Henderson-Hasselbalch equation.
pH = pKa + log(base)/(acid).
You CAN simply substitute 60 for base and 40 for acid and solve OR, if that is a little confusing, you can assume some convenient volume of acetic acid (say 50 mL or something like that), neutralize 60% of that, calculate base concn and acid concn and substitute. The fraction will always turn out to be 6/4 for the 60/40 ratio.
To determine the pH when 60% of the acetic acid has been neutralized, we need to find the concentration of acetic acid and its conjugate base at this point.
First, let's calculate the initial concentration of acetic acid (CH3COOH) in the solution. We are given that it is a 0.1 M solution.
Next, we need to determine the concentration of acetic acid remaining after 60% neutralization. Since 60% of the acid has been neutralized, 40% (100% - 60%) remains. We can calculate this by multiplying the initial concentration by 0.4:
Remaining concentration of CH3COOH = 0.1 M x 0.4 = 0.04 M
Now, we can find the concentration of the acetic acid's conjugate base, acetate ion (CH3COO-). Since acetic acid and acetate ion are in a 1:1 ratio in the dissociation reaction, the concentration of acetate ion will also be 0.04 M.
Using the equilibrium constant (Ka) for acetic acid, which is given as 1.8x10^-5, we can set up an equation to determine the concentration of H+ ions:
Ka = [H+][CH3COO-] / [CH3COOH]
Since we are interested in the pH, which is defined as the negative logarithm of the H+ ion concentration, we can rewrite the equation as:
pH = -log([H+])
We can rearrange the equation for the equilibrium constant to solve for [H+]:
[H+] = (Ka x [CH3COOH]) / [CH3COO-]
Plugging in the values, we get:
[H+] = (1.8x10^-5) x (0.04 M) / (0.04 M)
Simplifying,
[H+] = 1.8x10^-5
Finally, to get the pH, we take the negative logarithm of the [H+] concentration:
pH = -log(1.8x10^-5)
Calculating this value, we find:
pH ≈ 4.74
Therefore, the pH when 60% of the acetic acid has been neutralized is approximately 4.74.