PH question (volume calculation)

1. May 19, 2007

Endlicheri

I am having difficulties as to what equation to use to calculate the volume. Can someone please kindly give me some help?

1. The problem statement, all variables and given/known data

What is the volume of 6M HCl required to acidify the mixture of 1g benzaldehyde and 2mL 10M KOH to a pH ~2?

2. Relevant equations

MW of benzaldehyde is 106.1g/mol

cannizzaro reaction: benzaldehyde + KOH + H+ <=> benzoate + benzoic acid

pH = pKa + logQ <- I don't know if this is right to use....

3. The attempt at a solution
Here is what i think. The volume of HCl needed to acidify should be more than the volume needed to neutralize the solution, so I've calculated the amount needed for neutralization, which is according to my calculation:
Moles of Benzaldehyde
= 1g Benzaldehyde x (1 mol/106.1g)
= 0.009425 mol benzaldehyde
Mole of KOH
= (2 ml x 10 mmol/1ml) x (1 mol/1000 mmol)
= 0.02000 mol KOH
Mole of OH- in excess
= 0.02000 mol KOH – 0.009425 mol Benzaldehyde
= 0.01058 mol OH- in excess
To neutralize, volume of HCl needed
= 0.01058 mol HCl x (1 L HCl/ 6mol HCl) x (1000mL/1L)
= 1.76 mL HCl

But the problem is that I don't know how to relate the equation to the PH equation since KOH and H+ is both on the same side of the chemical equation.....

Last edited: May 20, 2007
2. May 21, 2007

chemisttree

The products of a Cannizzaro reaction of 2 moles of benzaldehyde and hydroxide are not benzoate and benzoic acid. You must correct that before you begin.