What mass of sodium benzoate should be added to 160.0mL of a 0.17M benzoic acid solution in order to obtain a buffer with a pH of 4.30?
mmols benzoic acid = mL x M = 160.0 x 0.17 = approx 27 but you need to do it more accurately.
pH = pKa + log (base)/(acid)
You will need to look up the pKa for benzoic acid.
Substitute pKa, (base) = x, (acid) = 27, pH = 4.30, and solve for x = benzoate
x = mmols benzoate.
Convert to grams sodium benzoate.
I tried working out the problem and was unable to get the correct answer. Could you possibly work it out entirely? I don't know what I'm doing wrong.
To determine the mass of sodium benzoate (NaC7H5O2) needed to make the desired buffer solution, we can use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
Where:
pH = desired pH of the buffer solution
pKa = -log(Ka), the acid dissociation constant of benzoic acid
[A-] = concentration of the conjugate base (sodium benzoate)
[HA] = concentration of the acid (benzoic acid)
First, we need to find the pKa value for benzoic acid:
The pKa value for benzoic acid is 4.20.
Next, we can rearrange the Henderson-Hasselbalch equation to solve for [A-]/[HA]:
[A-]/[HA] = 10^(pH - pKa)
Substituting the given values:
[A-]/[HA] = 10^(4.30 - 4.20)
[A-]/[HA] = 10^0.10
[A-]/[HA] = 1.26
This means that the concentration of the conjugate base (sodium benzoate) should be 1.26 times higher than the concentration of the acid (benzoic acid) in the buffer solution.
Since we are given the volume (160.0 mL) and concentration (0.17 M) of the benzoic acid solution, we can calculate the moles of benzoic acid (HA) using the formula:
moles = volume (L) x concentration (M)
moles HA = 0.160 L x 0.17 mol/L
moles HA = 0.0272 mol
Since the moles of benzoic acid and sodium benzoate should be equal in a buffer solution, the moles of sodium benzoate (A-) needed is also 0.0272 mol.
Finally, we can calculate the mass (g) of sodium benzoate using its molar mass (C7H5NaO2 = 144.11 g/mol) and the number of moles needed:
mass = moles x molar mass
mass sodium benzoate = 0.0272 mol x 144.11 g/mol
mass sodium benzoate = 3.92 g
Therefore, approximately 3.92 grams of sodium benzoate should be added to the 160.0 mL of the 0.17 M benzoic acid solution to obtain a buffer with a pH of 4.30.