The pH at 0.50 mL after the equivalence point is 11.14
Total V = 57.11 mL + 0.5 mL = 57.61 mL = 0.05761 L
nNaOH aggregate = 0.1532 mol / L * 0.05761 L = 8.83x10 ^ -3 moles
nHF = 8.75 x10 ^ -3 moles
nNaOH final = 8.83x10 ^ -3 moles - 8.75 x10 ^ -3 moles = 8x10 ^ -5 moles
[NaOH] final = [OH-] = 8x10 ^ -5 moles / 0.05761 L = 1.39x10 ^ -3 M
pOH = - Log (1.39x10 ^ -3) = 2.86
pH = 14 - 2.86 = 11.14.
Hence, pH at 0.50 mL after the equivalence point is 11.14
Learn more about pH here:
https://brainly.com/question/172153
#SPJ4
