[joey_h]

How to calculate the amount of 6M HCl that would be required to neutralize and acidify the reaction mixture of 1g benzaldehyde and 2mL 10M KOH to a pH of ~2?

Thanks

[zain]

How to calculate the amount of 6M HCl that would be required to neutralize and acidifythe reaction mixture of 1g benzaldehyde and 2mL KOH to a pH of 2?

Thanks

first write out a chemical equation. then think about the rest

[joey_h]

How to calculate the amount of 6M HCl that would be required to neutralize and acidifythe reaction mixture of 1g benzaldehyde and 2mL KOH to a pH of 2?

Thanks

first write out a chemical equation. then think about the rest

Can someone kindly teach me how to figure one?
The chemical equation i can think of is benzaldehyde + KOH + H+ => benzyl alcohol and benzoic acid
But what is the relationship between the H+ and the benzoic acid?

[refid]

Since KOH is used in x.s. you gotta neutralize all of it, then the only think i could think of is treat the benzoate  to benzoic acid as a buffer system using henderson hasselbach equation

[joey_h]

Since KOH is used in x.s. you gotta neutralize all of it, then the only think i could think of is treat the benzoate  to benzoic acid as a buffer system using henderson hasselbach equation

Here is what i think. please correct me if i am wrong

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

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

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..

[refid]

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
 
If you look at the mechanism half of the Benzaldehyde reacts with the OH- other half receives the aldehyde proton

So heres my guess after this step write the Volume u calculated

now calculate the amount to neutralize half the benzoate... that way we have a buffer system and can use the HH equation to solve the [A-]/[HA] ratio to have a pH~2. In turn, use this and the Ka to solve for [H+]. Then convert that to Volume of HCl, add to previous volume to neutral excess KOH.....I think..dont be mad at me is this wrong... just trying to help

refid