Background: If aqueous NaOH is added to tert-butyl chloride acetone solution, in

Question

Background: If aqueous NaOH is added to tert-butyl chloride acetone solution, initially the solution is basic due to the hydroxide ions from the NaOH. As the t-butyl chloride continues the reaction through hydrolysis and generates HCl, the HCl neutralizes the NaOH ions

Question:  If 5.0mL of 0.0091M aqueous NaOH is added to 5mL of 0.091M of tert-butyl chloride in acetone, calculate the pH at 10% hydrolysis.  (There is a hint listed that says to calculate moles of HCl produced and compare that number to the moles of NaOH present in the reaction mixture.)

Thank you for any help. If you could show work and provide a brief description, it would definitely be appreciated.

Explanation / Answer

millimoles of NaOH= .0455

millimoles of tertbutyl= .455

NaOH+ (CH3)3CCl ====> (CH3)3COH + HCl

Now, since one NaOH reacts with one t-butyl to give one HCl

10% hydrolysis of t-butyl means .0455 millimoles of t-butyl reacted using up .0455 millimoles of NaOH hence now no NaOh is left in the solution. This reaction will form .0455millimoles of HCl

since now volume of solution is 10ml

Molarity=.0455/10=.00455M

PH= 2.342

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