Acid-Base Reaction: Calculating PH of NaF Solution

Click For Summary

Discussion Overview

The discussion revolves around calculating the pH of a 0.020 mol/l sodium fluoride (NaF) solution formed after titrating sodium hydroxide with hydrofluoric acid. Participants explore the implications of the weak base nature of fluoride ions and the necessary equilibrium considerations for determining pH.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant states that the pH can be calculated directly from the concentration of fluoride ions, suggesting it is a weak base.
  • Another participant expresses confusion about calculating pH using the formula -log(0.020M), arguing that the solution should be basic due to the reaction of a strong base with a weak acid.
  • A later reply clarifies that the equilibrium concentration of fluoride ions (F-) is not simply 0.020M due to hydrolysis, and that Kb should be used to find the pH instead.
  • Participants discuss the need to derive Kb from Ka for hydrofluoric acid (HF) to proceed with the calculation.
  • There is a correction regarding the initial and equilibrium concentrations, with emphasis on the fact that the equilibrium concentration of HF must be calculated and is expected to be much smaller than 0.020M.
  • Some participants express uncertainty about the interpretation of the equilibrium concentrations and the implications for calculating pH.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the interpretation of the equilibrium concentrations and how they affect the calculation of pH. Multiple competing views remain regarding the correct approach to determining the pH of the solution.

Contextual Notes

There are unresolved assumptions regarding the initial and equilibrium concentrations of the species involved in the reaction, as well as the implications of the weak base behavior of fluoride ions on the pH calculation.

hallowon
Messages
35
Reaction score
0

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
 
Physics news on Phys.org
You look for pH of 0.020M F- solution, everything else doesn't matter. This is a weak base.

--
 
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
 
hallowon said:

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
 
Last edited:
thanks so much! you just made my day.
 
x12179x said:
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.

hallowon said:
thanks so much! you just made my day.

You will be badly surprsised if you assume x12179x approach.

--
 
Borek said:
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." ?
 
thats kinda what i assumed to
 
x12179x said:
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.

--
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 7 ·
Replies
7
Views
7K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 5 ·
Replies
5
Views
8K