Data Page Initial pH of acetic acid solution, pH -log(VKa THAD Calculations Mole
ID: 478744 • Letter: D
Question
Data Page Initial pH of acetic acid solution, pH -log(VKa THAD Calculations Moles of acetic acid: Molar mass of sodium acetate: Mass of sodium acetate needed: Calculations: Buffer capacity: Initial pH of 50.0 mL buffer solution: Calculations: Initial pH of 20.0 mL buffer solution: pH after addition of 0.100 M NaOH pH after addition of 0.100 M HCl Addition 1.00 mL added 1.00 mL added 2.00 mL added 2.00 mL added 3.00 ml, added 3.00 mL added 4.00 mL added 4.00 mL added 5.00 mL added 5.00 mL added CalculationsExplanation / Answer
moles of sodium acetate = 0.41/82.03 = 0.005 mols
moles of acetic acid = 0.005 moles
initial pH of buffer = pKa of acetic acid = 4.75
pH after addition of 0.1 M HCl to 50 ml buffer
(a) added 1 ml of HCl
molarity of sodium acetate = (5 - 0.1 x 1ml)mmol/51 ml = 0.0961 M
molarity of acetic acid = 5.1 mmol/51 ml = 0.1 M
pH = pKa + log(base/acid)
pH = 4.75 + log(0.0961/0.1) = 4.733
(b) added 2 ml of HCl
molarity of sodium acetate = (5 - 0.1 x 2ml)mmol/52 ml = 0.0923 M
molarity of acetic acid = 5.2 mmol/52 ml = 0.1 M
pH = pKa + log(base/acid)
pH = 4.75 + log(0.0923/0.1) = 4.715
Similarly other calculations can be done
pH after addition of 0.1 M NaOH to 50 ml buffer
(a) added 1 ml of NaOH
molarity of acetic acid = (5 - 0.1 x 1ml)mmol/51 ml = 0.0961 M
molarity of sodium acetate = 5.1 mmol/51 ml = 0.1 M
pH = pKa + log(base/acid)
pH = 4.75 + log(0.1/0.0961) = 4.767
(b) added 2 ml of NaOH
molarity of acetic acid = (5 - 0.1 x 2ml)mmol/52 ml = 0.0923 M
molarity of sodium acetate = 5.2 mmol/52 ml = 0.1 M
pH = pKa + log(base/acid)
pH = 4.75 + log(0.1/0.0923) = 4.785
Similarly other calculations can be done
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.