Acid-Base Reaction: Calculating PH of NaF Solution

[hallowon]

Homework Statement

after titrating sodium hydroxide with hydrofluoric acid, a chemist determined that the reaction had formed an aqueous solution of 0.020 mol/l sodium fluoride. determine ph of the solution

Homework Equations

The Attempt at a Solution

Ka of HF= 6.3X10^-4
NaOH + HF (equlibrium arrows) NaF +H20
I ? ? 0 0
C -X -X +X +X
E ?-X ?-X 0.020 0.020

ka = product/reactants i got stuckhere because there no initial value, and i don't know why I am trying to find initial when i need to find ph

 

[Borek]

You look for pH of 0.020M F^- solution, everything else doesn't matter. This is a weak base.

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[hallowon]

wait i don't get it ? -log 0.020M is 1.69 that is wrong b/c Strong base reacts with weak acid so ph should be aboe 7 b/c it will be a basic solution

 

[x12179x]

The Attempt at a Solution

Ka of HF= 6.3X10^-4
NaOH + HF (equlibrium arrows) NaF +H20
I ? ? 0 0
C -X -X +X +X
E ?-X ?-X 0.020 0.020

ka = product/reactants i got stuckhere because there no initial value, and i don't know why I am trying to find initial when i need to find ph

You don't need to know I, the initial, since we are given that the Equilibrium of NaF (C-X) is 0.020

The dissociation equation, as you said is:

NaF + H20 <==> NaOH + HF
(B + H2O <==> HB + OH-)

or:

F- + H2O <=> HF + OH-

I C ...N/A....
C C -X ... +X ... +X
E 0.020 ... +X ... +X

Since we have B + H2O <==> HB + OH-, the equation to use is then Kb=[HB][OH-]/[B-] so we need to get Kb from Ka so...

Kb = Kw/Ka = [10^-14]/[6.3x10^-4] = 1.58 x10^-11​

We now know:
Kb=1.58 x10^-11
[F-]=0.02
[OH-]=[HF]=X

All you have to do is plug these values into Kb=[HF][OH-]/[F-] and solve for X
Remember... X=[OH-], so you need to get the pH from the pOH Relevant equations:
B + H2O <==> HB + OH-
Kb = Kw/Ka
Kb=[HB][OH-]/[B-]
pOH= -log(OH-)
pOH + pH=14

 

[hallowon]

thanks so much! you just made my day.

 

[Borek]

You don't need to know I, the initial, since we are given that the Equilibrium of HF (C-X) is 0.20

No, we are given that initial F^- is 0.020M, and equilibrium concentration of HF is not 0.020M but has to be calculated. And in fact it is a lot smaller than 0.020M.

thanks so much! you just made my day.

You will be badly surprsised if you assume x12179x approach.

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[x12179x]

No, we are given that initial F^- is 0.020M, and equilibrium concentration of HF is not 0.020M but has to be calculated. And in fact it is a lot smaller than 0.020M.

Hmm, I mistyped HF instead of NaF above (and also .2 instead of .02); Now corrected

But I thought 0.02 was the equilibrium value of NaF since it said:

"after titrating...the reaction had formed an aqueous solution of 0.020 mol/l sodium fluoride." ?

 

[hallowon]

thats kinda what i assumed to

 

[Borek]

But I thought 0.02 was the equilibrium value of NaF since it said:

"after titrating...the reaction had formed an aqueous solution of 0.020 mol/l sodium fluoride." ?

Just because you have 0.020M solution of NaF doesn't mean 0.020M is an equilibrium concentration of F^-. Quite the opposite - F^- is a weak Broensted base, reacting with water, so its concentration after hydrolysis is smaller.

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