Question

Calculate the pH of a 0.30 M NaF solution. The Ka value for HF is 7.2*10^-4

Answer 1

Please explain the procedure involved in solving this kind of problem in general, medal's given if answer is clear

Answer 2

Lol we're not here competing for medals you know...

Answer 3

Totally right abb0t.
Okay, we look at the \(\sf F^{-}\) ion and remember it is a weak base. So we are going to look at the reaction:
\[\large \sf F^{-} + H _{2}O \to FH + OH ^{-}\]
The base dissociation constant become:
\[\large \sf K _{b}= \frac{ \left[ FH \right] \times \left[ OH ^{-} \right] }{ \left[ F ^{-} \right] }\]
We want to find the \(\sf H^{+}\) concentration so we need to find \(\sf \left[ OH^{-} \right]\). We do so by assuming that the reaction run \(x\) times and we then get:
\[\large \sf K _{b}=\frac{\it x \times x }{ \it 0.30-x }\]
Solve for \(x\) as \(x=\sf \left[ FH \right]=\left[ OH ^{-} \right]\)
When we got \(\sf \left[ OH ^{-} \right]\) we use the constant for the self-ionization of water which is equal to:
\[K _{w}=\sf \left[ OH ^{-} \right] \times \left[ H ^{+} \right]=10^{-14} ~ M ^{2} ~ ~ ~ ~ T=25^{\circ}C\]
Then use the definition of pH to find pH.

Answer 4

Oh yea forgot:
Start by calculating \(\sf K_{b}\).

Answer 5

@Frostbite , The value of Ka is given! so I think there is no need to calculate the value of Kb. And our Chem teacher mentioned something called " 5% rule " that can be used to discard the "-x" part in "0.30-x". How does that comes in?

Answer 6

sorry it should be HF solution, not NaF. I misread the question last night because I was too tired

Answer 7

Well, I just answered what you asked. :)
About the "5% rule" I don't know what that is. We do have ways to approximate our self through acid/base calculations, but usually it is just better not to do so, as it is more easy and you don't do it wrong.

Answer 8

A buffer solution may be acidic or basic. Your question involves the acidic buffer. An acidic buffer solution contains a weak acid and its salt.
In your question the weak acid is hydrofluoric acid;
HF .... Ka = 7.2x10^-4
and its salt is sodium fluoride (NaF)

As these formulas imply hyrdofluoric acid is a monoprotic acid.

Concentrations of the substances;
[HF] = 0.20 M
[NaF] = 0.10 M

(a) Dissociations:
Weak acid ionizes only a small extent.
HF(aq) <-----> H+(aq) + F-(aq)
0.20 - x M ...... x M ........ x M

Ionic soluble salt ionizes completely.
NaF(aq) -----> Na+(aq) + F-(aq)
0.10 M ........... 0.10 M ..... 0.10 M

F- is the common ion, but the concentration of F- ion produced by the dissociation of the salt (NaF) will be very large compared to the concentration of F- ion produced by the dissociation of the acid, and hence x is neglected.

Ka = [H+][F-] / [HF]

This expression in some cases simplified as (since x is neglected)

Ka = [H+][SALT] / [ACID]

Substituting the values;

7.2x10^-4 = [H+] (0.10) / (0.20)
[H+] = (7.2x10^-4 x 0.20) / 0.10 = 1.44x10^-3

Awnser:1.4 x 10-3 M

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