What CH3COONa concentration is required to prepare a buffer solution with a pH of 4.80 if [CH3COOH] = 0.150 M? Ka of CH3COOH = 1.8 10-5.
If I use your numbers I come out with 1.7E-4. The difference arises from 0.001135*0.15 = 1.7E-4 and I assume you just punched a wrong button; however, none of that counts because you didn't use the right Ka.
Ka = 1.8E-5 so pKa = 4.74
sorry read it wrong, doing this now in chem
To find the concentration of CH3COONa required to prepare a buffer solution, you can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
In this case, the acid is CH3COOH, and its conjugate base is CH3COO-. The pKa value can be calculated using the Ka value given:
pKa = -log(Ka) = -log(1.8 × 10^-5) = 4.74
Substituting the given values into the Henderson-Hasselbalch equation, we have:
4.80 = 4.74 + log([CH3COO-]/[CH3COOH])
Next, we rearrange the equation to isolate the ratio [CH3COO-]/[CH3COOH]:
log([CH3COO-]/[CH3COOH]) = 4.80 - 4.74 = 0.06
Now, we can convert the logarithmic equation into an exponential equation:
[CH3COO-]/[CH3COOH] = 10^(0.06)
Solving for [CH3COO-]/[CH3COOH], we find:
[CH3COO-]/[CH3COOH] ≈ 1.15
Since the concentration of CH3COOH is given as 0.150 M, we can calculate the concentration of CH3COO-:
[CH3COO-] = 1.15 × 0.150 M ≈ 0.173 M
Therefore, the required concentration of CH3COONa (which is the source of CH3COO-) to prepare a buffer solution with a pH of 4.80 is approximately 0.173 M.
4.80 = pKa + log(x)/(0/150)
Substitute pKa and solve for x.
PH=-log(ka)+log(base/acid)
4.8=-log(1.8*10^-8)+log(x/.15)
4.8=7.7447+log(x/.15)
-2.9447=log(x/.15)
.001135=x/.15
x=1.875*10^-4