Question

An aqueous solution contains 0.390 M hydrofluoric acid. How many mL of 0.200 M sodium hydroxide would have to be added to 250 mL of this solution in order to prepare a buffer with a pH of 3.050? mL

if you need a value of Ka, you should use this table:

Answer 1

Calculating pka: Given Ka = 7.2x10-4 pka = - log [ Ka ] pka = - log [ 7.2x104 1 | pka = 3.14 | number of moles = Molarity * Volume volume of HF = 250 mL= 0.250 L initial no. of moles of HF = 0.390 X 0.250 = 0.098 mol Construct ICA table : Reaction HF + NaOH → Naf concentration ( mol ) 0.098 x change ( mol ) -X - -x xx equilibrium ( mol ) 0.098-X 0

Calculating moles: According to Henderson-Hasselbalch equation [ salt pH = pka + log I weak acid 1 pH = pka + log [ NaF ] [ HF ] [ X ] [ 0.098-X ] 3.05 = 3.14 + log = 0.808 [ 0.098-X ] 1.808 x = 0.079 NaOH moles = X = 0.044 Volume = Mole / Molarity Molarity of NaOH = 0.20 M. Volume of NaOH = 0.044 / 0.20 = 0.218 L | Volume of NaOH = 218 ml