I am having difficulties as to what equation to use to calculate the volume. Can someone please kindly give me some help?
What is the volume of 6M HCl required to acidify the mixture of 1g benzaldehyde and 2mL 10M KOH to a pH ~2?
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....
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.....