How many moles of sodium acetate
(NaCH3COO) must be added to 1.000 liter of a 0.500M solution of acetic acid (CH3COOH) to produce a pH of 5.027? The ionization constant of acetic acid is 1.8 × 10−5.
Answer in units of mol
To answer this question, we need to understand the concept of the Henderson-Hasselbalch equation and how pH relates to the concentration of a weak acid.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
where pH is the desired pH, pKa is the negative logarithm of the acid dissociation constant (Ka) of the weak acid, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.
In this case, acetic acid (CH3COOH) is a weak acid, and its conjugate base is acetate (CH3COO-).
We are given the pH as 5.027 and the ionization constant (Ka) of acetic acid as 1.8 × 10^-5.
1. Calculate the pKa using the given Ka:
pKa = -log(Ka) = -log(1.8 × 10^-5)
2. Substitute the given pH, pKa, and [A-] into the Henderson-Hasselbalch equation to find [HA]:
pH = pKa + log([A-]/[HA])
5.027 = pKa + log([A-]/[HA])
3. Rearrange the equation to isolate [HA]:
log([A-]/[HA]) = 5.027 - pKa
4. Take the inverse logarithm of both sides to remove the logarithm:
[A-]/[HA] = 10^(5.027 - pKa)
5. Since [A-] = [NaCH3COO] and [HA] = [CH3COOH], we can substitute these values:
[NaCH3COO]/[CH3COOH] = 10^(5.027 - pKa)
6. Convert the given 0.500M acetic acid solution to moles per liter:
[CH3COOH] = 0.500 moles/liter
7. Substitute the pKa value and the concentration values into the equation and solve for [NaCH3COO]:
[NaCH3COO]/0.500 = 10^(5.027 - pKa)
By rearranging and solving the equation, you can calculate the concentration of sodium acetate ([NaCH3COO]). Multiply this concentration by the total volume of the solution (1.000 liter) to obtain the number of moles of sodium acetate needed.
Use the HH equation and solve for Ac.
pH = pKa + log (Ac/HAc)
5.027 = 4.744 + log Ac/0.5
Now convert M Ac to grams.