PH question (volume calculation)

[Endlicheri]

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

Homework Statement

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

Homework 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...

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...

 

[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.

 

[Blueshift5]

I can understand your confusion in trying to determine the appropriate equation to use to calculate the volume in this situation. The equation you have listed, pH = pKa + logQ, is typically used to calculate the pH of a solution based on the concentration of the acid and its dissociation constant. In this case, you are trying to determine the volume of a strong acid (HCl) needed to reach a specific pH, rather than calculating the pH itself.

To solve this problem, you can use the concept of stoichiometry, which involves using the balanced chemical equation to determine the relationship between the reactants and products. In this case, the balanced equation for the reaction between benzaldehyde and KOH is:

benzaldehyde + KOH + H+ <=> benzoate + benzoic acid

From this equation, we can see that for every 1 mole of benzaldehyde, we need 1 mole of KOH and 1 mole of H+ to produce 1 mole of benzoate and 1 mole of benzoic acid.

In your calculation, you correctly determined the moles of benzaldehyde and KOH present in the mixture. However, to determine the volume of HCl needed, we need to consider the molar ratio between H+ and KOH. Since the concentration of KOH is 10M, we know that for every 1 mole of KOH, there is 1 mole of H+. Therefore, the moles of H+ needed to neutralize the mixture is also 0.02000 mol.

To calculate the volume of HCl needed, we can use the following equation:

Volume of HCl = (Moles of H+ needed) x (Molarity of HCl) x (Volume of solution)

= (0.02000 mol H+) x (6 mol HCl/1 mol H+) x (2 mL solution)

= 0.24 mL HCl

Therefore, the volume of 6M HCl needed to acidify the mixture to a pH ~2 is 0.24 mL. I hope this helps clarify the approach to use in this type of problem.