Question

Calculate the pH of the resulting solution formed by adding 10.00mL of 0.200M NaOH to 20.00mL of 0.200 M HF K_a=6.8x10^-4 assume small x approximation is okay.

Answer 1

we have:

Molarity of HF = 0.2 M

Volume of HF = 20 mL

Molarity of NaOH = 0.2 M

Volume of NaOH = 10 mL

mol of HF = Molarity of HF * Volume of HF

mol of HF = 0.2 M * 20 mL = 4 mmol

mol of NaOH = Molarity of NaOH * Volume of NaOH

mol of NaOH = 0.2 M * 10 mL = 2 mmol

We have:

mol of HF = 4 mmol

mol of NaOH = 2 mmol

2 mmol of both will react

excess HF remaining = 2 mmol

Volume of Solution = 20 + 10 = 30 mL

[HF] = 2 mmol/30 mL = 0.0667M

[F-] = 2/30 = 0.0667M

They form acidic buffer

acid is HF

conjugate base is F-

Ka = 6.8*10^-4

pKa = - log (Ka)

= - log(6.8*10^-4)

= 3.167

we have below equation to be used:

This is Henderson–Hasselbalch equation

pH = pKa + log {[conjugate base]/[acid]}

= 3.167+ log {6.667*10^-2/6.667*10^-2}

= 3.167

Answer: 3.17