Acetic acid (CH3COOH) has a Ka of 1.8x10^-5. a buffer is to be prepared with a pH of 5.00 from solutions of CH3COOH and sodium acetate (NaCH3COO) of the same concentration. How many mLs of NaCH3COO would have been added to 100 mL of CH3COOH to make the proper buffer solution??
Use the Henderson-Hasselbalch equation.
5.00 = 4.76 (or whatever you have for pKa) + log(base)/(acid)
Solve for B/A.
You don't have a molarity of either given BUT you can make it easy and assign 1.0M to each.
I did a quickie calculation and came up with about 174 mL but you should confirm that and do it more accurately.
To solve this problem, we need to use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the pKa and concentration of the acid and its conjugate base.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
Where:
pH is the desired pH of the buffer (5.00 in this case)
pKa is the dissociation constant of the acid (1.8x10^-5 in this case)
[A-] is the concentration of the conjugate base (NaCH3COO)
[HA] is the concentration of the acid (CH3COOH)
We are given that the concentration of CH3COOH and NaCH3COO are the same, so let's call it C.
Now we can rewrite the Henderson-Hasselbalch equation as:
5.00 = -log(1.8x10^-5) + log([NaCH3COO]/[CH3COOH])
Simplifying, we have:
5.00 = 4.74 + log([NaCH3COO]/[CH3COOH])
Subtracting 4.74 from both sides:
0.26 = log([NaCH3COO]/[CH3COOH])
To eliminate the logarithm, we can take the inverse logarithm (antilog) of both sides:
10^0.26 = [NaCH3COO]/[CH3COOH]
Now we can solve for [NaCH3COO]/[CH3COOH]:
[NaCH3COO]/[CH3COOH] = 10^0.26
Now we know that the ratio of [NaCH3COO]/[CH3COOH] is equal to 10^0.26. Since the concentration of CH3COOH is the same in both solutions, we can set it to be 100 mL (just to keep the calculations consistent).
[NaCH3COO] = 10^0.26 * [CH3COOH]
[NaCH3COO] = 10^0.26 * 100 mL
[NaCH3COO] = 1.81 * 100 mL
[NaCH3COO] = 181 mL
Therefore, to make the proper buffer solution, you would need to add 181 mL of NaCH3COO to 100 mL of CH3COOH.