Consider the titration of 44.4 mL of 0.250 M HF with 0.220 M NaOH. Calculate the
ID: 761009 • Letter: C
Question
Consider the titration of 44.4 mL of 0.250 M HF with 0.220 M NaOH. Calculate the pH after the addition of 11.1mL of baseExplanation / Answer
HF(aq) --> H+(aq) + F-(aq) Ka = 7.08 * 10^-4 you add 10 ml 0.2 M NaOH [HF] initial = 0.25 mole/L = x moles/ 0.040 L ==> x moles = 0.01 moles HF 10ml * 0.2 moles NaOH/1000ml = 0.002 moles OH- by adding the base you deplete the HF but you add F- so you are now building a buffer mole HF left = 0.01 - 0.002 moles = 0.008 moles HF moles F- generated = 0.002 final volume = 40 ml + 10 ml = 50 ml = 0.050 L [HF] final = 0.008 moles HF/0.050 L = 0.16 M [F-] final = 0.002 moles F-/0.050 L = 0.04 M pH = pKa + log[F-]/[HF] = 3.15 + log(0.04/0.16) = 2.55