Question

A sample of 0.250 M hydrofluoric acid (HF) is reacted with 0.150 M sodium hydroxide (NaOH) solution. Ka = 6.8 x 10-4 for HF. What is the pH when you mix 25.0 mL of HF with 50.0 mL of NaOH.

Answer 1

Given:

M(HF) = 0.25 M

V(HF) = 25 mL

M(NaOH) = 0.15 M

V(NaOH) = 50 mL

mol(HF) = M(HF) * V(HF)

mol(HF) = 0.25 M * 25 mL = 6.25 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.15 M * 50 mL = 7.5 mmol

We have:

mol(HF) = 6.25 mmol

mol(NaOH) = 7.5 mmol

6.25 mmol of both will react

excess NaOH remaining = 1.25 mmol

Volume of Solution = 25 + 50 = 75 mL

[OH-] = 1.25 mmol/75 mL = 0.0167 M

use:

pOH = -log [OH-]

= -log (1.667*10^-2)

= 1.7782

use:

PH = 14 - pOH

= 14 - 1.7782

= 12.2218

Answer: 12.22