Calculate the pH of a mixture of one mole of benzoic acid and two mole of sodium benzoate dissolved in 1 L of solution. The pKa of benzoic acid is 4.2. What is the final molar concentration of the buffer?

pH = pKa + log (base)/(acid)

(acid) = mols benzoic acid/1 L = ?
(base) = mols sodium benzoate/1 L = ?
Substitute and solve for pH.

Final concentration of buffer = total mols/total L = mols acid + mols base/total mols = 3 mol/L = ?

To calculate the pH of a mixture of benzoic acid and sodium benzoate, we need to consider the Henderson-Hasselbalch equation, which is given by:

pH = pKa + log([A-]/[HA])

Where pH is the desired pH, pKa is the dissociation constant of the acid (benzoic acid in this case), [A-] is the concentration of the conjugate base (sodium benzoate), and [HA] is the concentration of the acid (benzoic acid).

In this case, one mole of benzoic acid and two moles of sodium benzoate are dissolved in 1 L of solution. The moles of sodium benzoate can be converted to concentration by dividing by the volume of the solution, which is 1 L.

The molar concentration of sodium benzoate ([A-]) = 2 moles / 1 L = 2 M

Since benzoic acid and sodium benzoate are in a 1:2 mole ratio, the molar concentration of benzoic acid ([HA]) = 1 mole / 1 L = 1 M

Now we can substitute these values into the Henderson-Hasselbalch equation:

pH = 4.2 + log(2/1)

Calculating this expression gives:

pH = 4.2 + log(2) = 4.2 + 0.3010 = 4.5010

Therefore, the pH of the mixture is approximately 4.5.

To determine the final molar concentration of the buffer, we can either use the concentration of the acid ([HA]) or the concentration of the conjugate base ([A-]), as they are in a 1:2 ratio.

In this case, we can use the concentration of the conjugate base:

Molar concentration of the buffer = [A-] = 2 M

So, the final molar concentration of the buffer is 2 M.