Can you please show detailed steps on how the correct answer was obtained? Calcu
ID: 637848 • Letter: C
Question
Can you please show detailed steps on how the correct answer was obtained?
Calculate the pH during the titration of 20.00 mL of 0.1000 M CH3COOH(aq) with 0.1000 M NaOH(aq) after 20.3 mL of the base have been added. Ka of acetic acid = 1.8 x 10-5.
Can you please show detailed steps on how the correct answer was obtained?
Calculate the pH during the titration of 20.00 mL of 0.1000 M CH3COOH(aq) with 0.1000 M NaOH(aq) after 20.3 mL of the base have been added. Ka of acetic acid = 1.8 x 10-5.
Selected Answer:6.56
Correct Answer:10.87 ± 0.02
Explanation / Answer
Moles of CH3COOH = volume x molarity
= 20 mL x 1L/1000 mL x 0.100 mol/L
= 0.00200 mol
Moles of NaOH = molarity x volume
= 20.3 mL x 1L/1000 mL x 0.100 mol/L
= 0.00203 mol
Total volume = 20 + 20.3 = 40.3 mL = 0.0403 L
NaOH + CH3COOH = CH3COONa + H2O
Moles of OH- in excess = 0.00203 - 0.00200
= 0.00003 mol
[OH-] = moles in excess / total volume
= 0.00003 mol / 0.0403 L
= 0.000744 M
pOH = - log [OH-] = - log (0.000744) = 3.128
pH = 14 - pOH = 14 - 3.128
pH = 10.872