A student does a titration to make an ester out of ethanol acid and ethanol. The student adds 1.0 moldm-3 of NaOH to the reaction mixture in a flask.But the NaOH actually used was more concentrated than 1.0 moldm-3. What would be his effect on the kc value?
Here's the equation--> CH3COOH+CH3CH2OH--> CH3COOCH3 +H20
My teacher said that he/she adds less NaOH and thinks there is less acid in the mixture therefore less reactants more products so Kc increases.
I don't really understand this. Shouldn't the equilibrium shift left in the more acidic part? What would be the effect if the NaOH was actually less concentrated?
Please could you explain this concept to me. I have a graded assement on this tomorrow and really need to grasp this. Thank you!