Question

Consider the titration ofa 25.0 mL solution0.150 M HF with 0.150 M NaOH. What is the pH of the solution before addition of titrant, after addition of 10.0mL, and after addition of 25.0mL of titrant?

Answer 1

1. Since HF is a weak acid. So we need ICE table to get the initial pH.

HF + H2O H3O+ + F-

Initial 0.150 0 0

Change -x +x +x

Equilibrium 0.150-x +x +x

Now substitute values into ka expression and solve for x

x² / 0.150-x = 6.6*10^-4

x² + 6.6*10^-4 x - 0.99*10^-4 = 0

x = -6.6*10^-4+ (6.6*10^-4)² - 4(1) (-0.99*10^-4) / 2

we get x = 0.96253*10^-2

So pH = -log0.96253*10^-2 = 2.02

Thus, initial pH = 2.02

2. After adding 10 mL of 0.150M NaOH

The number of mmoles of HF = 25 mL*(0.150mmol HF/1mL) = 3.75mmol HF

The number of mmoles of OH- added = 10mL*(0.150mmol OH-/1mL) = 1.5mmol OH-

To calculate pH, construct ICE table

HF + OH- H2O + F-

initial 3.75 0 0

add 0 1.5   

change -1.5 -1.5 1.5

equilibrium 2.25 0 1.5

To get the concentration, we must divide mmoles by total volume.

total volume = 25mLHF + 10 mL NaOH = 35 mL

concentration of HF = 2.25mmol/35mL = 0.0643M

concentration of F^-= 1.5mmol/35mL = 0.0429 M

Since an acid and its conjugate base are in equilibrium , we can use Henderson--hasselbalch equation to get the pH.

pH = pKa + log [F-]/[HF] = -log(6.6*10^-4) + log 0.0429/0.0643 = 3.00

3. After adding 25 mL 0f 0.150M NaOH

mmoles of OH^- added = 25.0mL * 0.150mmol/mL = 3.75mmol OH^-

now construct ICE table as follows:

HF + H2O H3O+ + F-

initial 3.75 0 - 0

add 0 3.75 - 3.75

change -3.75 -3.75 3.75

equiibrium 0 0 3.75

total volume = 25mL + 25mL = 50 mL

concentration of F^- = 3.75mmol / 50mL = 0.075 M

This is equivalence point of the titration. acid and base are both neutralized and neither is in excess.

However, to get the pH at this point, we must realize that F- will hydrolyze.

construct a ICE table

HF + H2O H3O+ + F-

initial 0.075 - - 0

change -x - +x +x

equilibrium 0.075-x x x

In this reaction F- acts as a base. So, we must get the kb value.

kb = kw / ka

kb = 1.0*10^-14 / 6.6*10^-4 = 1.515*10^-11

so, now 1.515*10^-11 = x2 / 0.075-x

solve for x, we get

x= 1.0659*10^-6

here we get pOH value as pOH = -log(1.0659*10^-6) = 5.97

Thus, pH = 14 - 5.97 = 8.03