Solution D: 20.0 mL of 1:1 buffer5.0 mL NaOH Save&Continue; Show how you calcula
ID: 559809 • Letter: S
Question
Solution D: 20.0 mL of 1:1 buffer5.0 mL NaOH Save&Continue; Show how you calculated the stoichiometric concentrations of HAc and Ac after mixing the buffer and base (ICE Table and any calculations) How many moles of NaOH are there in Solution D before the reaction takes place? Think about the concentration and the volume of NaOH How many moles of Acetic Acid are there in Solution D before the reaction takes place? Think about the concentration of acetic acid in the buffer and the volume of buffer used How many moles of Acetate Ion are there in Solution D before the reaction takes place? Think about the concentration of acetate ion in the buffer and the volume of buffer used Show a complete calculation of how you calculated the pH after mixing Use the pKa of acetic acid that you determined from the buffer above in your calculation Calculated pH of solution DExplanation / Answer
Calculation of pH of the buffer solution after the addition of base NaOH
pH of weak acid HAc and conjugate base Ac-
let 1 M be the concentration of HAc and Ac- in the buffer. Also 1 M NaOH is added
then,
initial moles of acetic acid in solution before addition of base = 1 M x 20 ml = 20 mmol
initial moles of acetate in solution before addition of base = 1 M x 20 ml = 20 mmol
moles of NaOH in solution D before addition of base = 0 mmol
moles of NaOH added = 1 M x 5 ml = 5 mmol
So,
ICE chart of HAc/Ac- reacting with base
HAc + OH- (base) <==> Ac- + H2O
Initial 20 5 20 -
Change -5 -5 +5 +5
Equilibrium 15 0 25 5
Thus,
using Hendersen-hasselbalck equation for solution D,
pH = pKa + log([Ac-]/[HAc])
pKa of acetic acid = 4.75
So,
pH = 4.75 + log(25/15)
= 4.972
The pH of solution D after addition of base NaOH = 4.972
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