You are titrating 50.00 mL of a 0.100 M HF (hydrofluoric add) solution with 0.10

Question

You are titrating 50.00 mL of a 0.100 M HF (hydrofluoric add) solution with 0.100 M KOH. The pK_a of HF is 3.14. HF (aq) + OH^- (aq) reversible F^- (aq) + H_2O (l) Once you have added 50.00 mL of KOH solution: What major* species are still present in solution (mark all that apply) HF OH^- F^- What is the pH of the solution acidic neutral basic *Only mark species that will be present after the titration reaction has finished, but before taking into account any small amounts formed at equilibrium or from the auto-ionization of Water. Assuming you have already taken into account any dilutions from the titration, what method would you use to solve for the actual pH Use the excess OH^- to calculate the pH Use the Henderson-Hasselbalch equation Use an ICE table and the K_a Use an ICE table and the K_b

Explanation / Answer

Given:

Volume of HF = 50.00 mL

Molarity of HF = 0.100 M

Molarity of KOH = 0.100 M

Pka of acid = 3.14

Volume of KOH = 50.00 mL

Reaction

HF (aq) + OH- (aq) -- > F- (aq) + H2O (l)

From this reaction we say that mole ratio of acid and base is 1: 1

At equivalence point both acid and base will have equal moles.

From given data it is predicted that volume and molarity of both are same. So moles of both are having equal moles.

Lets calculate moles

Moles of acid = volume in L x molarity = 0.050 L x 0.100 M = 0.0050 mol = moles of base

Now from the reaction we also say that moles F- formed will be equal in ration with acid base.

So at equilibrium moles of F- = 0.005 mol

Lets find out concentration of F-

[F-] = mol / Total volume in L

=[ 0.005 / (0.05+0.05) ] M

= 0.05 M

We know F- is conjugate base weak acid so it will again react with H2O. Lets show its reaction.

Reaction of F- with H2O and ICE

F-    +        H2O ----- ->      HF + OH-

I           0.05                               0            0

C          -x                                    +x                   +x

E          (0.05-x)                                       x          x

Now we write kb expression for this reaction.

Kb = [ HF ] [OH- ] / [ F-]

We can get kb value by using ka

First lets find out ka by using pka

Ka = Antilog (-pka)

=7.24 E-4

Kb = 1.0 E-14 / ka

= 1.0 E-14 / 7.24 E-4

=1.38 E-11

Lets plug this value in order to find x

1.38 E-11 = x2 / ( 0.05 – x)

The value of kb is very small that we can use 5 % approximation and so we neglect x value in denominator.

1.38 E-11 = x2 / 0.05

Now we find for x

x = 8.30 E-7 M

x = [OH-] = 8.30 E-7 M

Lets find out pOH

pOH = - log [OH-]

= - log (8.30 E-7 )

= 6.08

Now we can get pH by using following reaction.

pH = 14 – pOH

= 14 – 6.08 = 7.92

So the pH of the solution is 7.9

Question . 1st :

When it solution is neutralized there is F-

2)

pH of the solution   = 7.92

3).

If we use dilution method then we could have used Henderson Hasselbalch equation.

In this equation, At half equivalent point the pH or pOH equal to ka or kb respectively.

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