you want to make a 0.5 m acetate buffer at a ph = 5.5. how much acetic acid and base should you mix
Your prof PROBABLY wants you to add NaOH to HAc to prepare but you didn't show molarity of acids or bases available so I didn't do it that way.
To calculate the amounts of acetic acid and sodium acetate (base) needed to make a 0.5 M acetate buffer at pH 5.5, you'll need to use the Henderson-Hasselbalch equation. This equation relates the pH, pKa (the acid dissociation constant of the acid), and the ratio of the concentrations of the acid and its conjugate base:
pH = pKa + log([A-]/[HA]),
Where [A-] represents the concentration of the conjugate base and [HA] represents the concentration of the acid.
1. Determine the pKa value for acetic acid. The pKa of acetic acid is 4.76.
2. Calculate the ratio of [A-]/[HA] using the Henderson-Hasselbalch equation. Rearranging the equation gives us:
log([A-]/[HA]) = pH - pKa.
Using the given values, the equation becomes:
log([A-]/[HA]) = 5.5 - 4.76 = 0.74.
Taking the antilog (inverse logarithm) of both sides of the equation gives us:
[A-]/[HA] = 10^0.74 = 5.01187,
Therefore, the ratio of [A-]/[HA] is approximately 5.01187.
3. Determine the total concentration of the buffer solution. In this case, the desired concentration is 0.5 M.
4. Set up a system of equations to calculate the concentrations of acetic acid and sodium acetate.
Let x be the concentration of acetic acid (HA) and 5.01187x be the concentration of sodium acetate (A-).
The equation can be written as:
x + 5.01187x = 0.5,
6.01187x = 0.5,
x ≈ 0.083 M.
Therefore, you will need approximately 0.083 M of acetic acid and 5.01187 * 0.083 M = 0.416 M of sodium acetate (base) to make a 0.5 M acetate buffer at pH 5.5.
Do you mean 0.5 m or 0.5M. I see you don't seem to know there is a caps key on the keyboard. There is and I suggest you find it. m means molality and M means molarity. I assume you meant M. The next assumption is what volume do you want. Let's say 1000 mL.
pH = pKa + log (Ac^-)/(HAc)
Look up pKa, substitute into the equation, and solve for the ratio of (Ac^-)/(HAc). That is equation 1.
Eqquation 2 is HAc + Ac^- = 0.5.
Solve the two equation simultaneously for (HAc) and (.5)
Now, knowing (HAc) and (Ac^-) and that you want a liter of solution, calculate grams HAc and Ac^-.
Post your work if you get stuck.