Question

Determine the pH at the equivalence point for the titration of 25.0 mL of 0.50M HF solution with 0.40 M NaOH solution. This is what I have so far: First determine Vb = Veq Ma * Va = Mb * Vb 0.5 * 25 = 0.4 * Vb Vb = 31.25 mL Moles of HF = 0.5 M * 0.025 L = 0.0125 mol HF Moles of NaOH = 0.4 M * 0.03125 L = 0.0125 mol NaOH Total Volume = 0.03125 + 0.025 = 0.05625 L Molarity of base = 0.0125 / 0.05625 = 0.222 M OH-

Answer 1

at equivalence point

Ma x Va = Mb x Vb

so

0.5 x 25 = 0.4 x Vb

Vb = 31.25 ml

now

we know that

moles = concentration x volume (L)

so

moles of HF = 0.5 x 25 x 10-3 = 0.0125

moles of NaOH = 0.4 x 0.03125 = 0.0125

now

total volume = 25 + 31.25 = 56.25 ml = 0.05625 L

now

the reaction is

NaOH + HF --> NaF + H20

moles of NaF formed = moles of HF reacted = 0.0125

now

[NaF ] = moles / volume

[NaF] = 0.0125 / 0.05625

[NaF] = 0.2222

now

we know that

NaF is a weak base

for weak bases

[OH-] = sqrt ( Kb x C)

also

Kb = 10-14 / Ka

so

[OH-] = sqrt ( 10-14 x C / Ka)

we know that

Ka for HF is 6.76 x 10-4

so

[OH-] = sqrt( 10-14 x 0.222 / 6.76 x 10-4)

[OH-] = 1.813 x 10-6

we know that

pOH = -log [OH-]

so

pOH = -log 1.813 x 10-6

pOH = 5.74

now

pH = 14 - pOH

so

pH = 14 - 5.74

pH = 8.26

so

pH at equivalence point is 8.26