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PH question (volume calculation) 
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#1
May1907, 11:11 PM

P: 4

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..... 


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