PH of 6,0*10^-4 M NaNO2 solution

[TheSodesa]

Homework Statement

It's all in the title. Correct answer: pH = 7,20. [itex]K_a (HNO_2) = 4,0 \cdot 10^{-4}[/itex]

Homework Equations

[tex]K_b = \dfrac{K_w}{K_a}[/tex]
[tex]K_b = \dfrac{HB^+ \cdot OH^-}{B}[/tex]

The Attempt at a Solution

NaNO2 fully dissolves in water giving [itex]6,0 \cdot 10^{-4} M \, NO2^-[/itex]. Now water forms an equilibrium with NO2- as it gives H+ to NO2-:

[tex]NO_2 ^- + H_2 O \leftrightharpoons HNO_2 + OH^-[/tex]

[tex]
\begin{bmatrix}
c & NO_2^- & HNO_2 & OH^-\\
\hline
At \, start & 6 \cdot 10^{-4} & 0 & 0 \\
At \, equilibrium & 6 \cdot 10^{-4}-x & x & x
\end{bmatrix}
[/tex]

Now

[itex]K_b = \dfrac{x^2}{6 \cdot 10^{-4}-x} \leftrightarrow x^2 + K_b x - 6K_b \cdot 10^{-4} = 0 \\

\rightarrow x = \dfrac{-K_b \stackrel{+}{(-)} \sqrt{(K_b)^2 - 4(1)(-6K_b \cdot 10^{-4})}}{2}
= 1,22462 \cdot 10^{-7} = [HO^-]\\

\rightarrow pH = 14 - pOH = 14 + lg[OH^-] = 7,088
[/itex]

This is not the correct answer (7,20). What am I missing? Water? I only know the iterative process for determining the correct [H+] for low concentrations H+ (when water has an effect on the result as Zumdahl describes it), not OH-.

Can someone point me in the right direction?

 

[Borek]

Looks like it is one of these cases where water autodissociation can't be ignored. 7.20 is an answer I got for an exact solution (done with BATE, pH calculator), 7.09 is what the ICE table produces.

In general this requires solving a 3^rd degree polynomial.

 

[TheSodesa]

Looks like it is one of these cases where water autodissociation can't be ignored. 7.20 is an answer I got for an exact solution (done with BATE, pH calculator), 7.09 is what the ICE table produces.

In general this requires solving a 3^rd degree polynomial.

Ok, now I'm baffled, since unless the polynomial is in product form (=0) I don't think I know how to solve such a thing. So the formula [tex]K_a = \dfrac{[H^+]^2 - K_w}{[HA]_0 - \dfrac{[H^+]^2 - K_w}{[H^+]}}[/tex] isn't going to help me here? What I thought of doing is finding out the [H+] from the calculated pH and using the above formula, but it would require knowing [itex] [HA]_0 [/itex], meaning the amount of acid put into the solution before it dissociated. Sadly that is information I don't have (and there was no acid to begin with, just a salt of an acid).

 

[Borek]

Assuming NaOH (more precisely: Na⁺) can be ignored and the only source of pH change is the base NO₂^- you can follow approach described here:

http://www.chembuddy.com/?left=pH-calculation&right=pH-acid-base-solution

You know the total amount of acid, why do you say you don't know it? It is what is given.

 

[TheSodesa]

Assuming NaOH (more precisely: Na⁺) can be ignored and the only source of pH change is the base NO₂^- you can follow approach described here:

http://www.chembuddy.com/?left=pH-calculation&right=pH-acid-base-solution

You know the total amount of acid, why do you say you don't know it? It is what is given.

Turns out it was much simpler and I didn't have to go beyond what the book teaches. I tried the iteration process to find out [OH^-], but with K_b, [ B ]₀ and [OH^-] instead of the values used in the formula for acids in the book as follows

[tex] K_b = \dfrac{[OH^-]^2 - K_w}{[ B ]_0 - \dfrac{[OH^-]_0^2 - K_w}{[OH^-]_0}} [/tex]

, and got the right answer. I suppose I could have derived the formula for bases myself by applying mass-, charge- and other conservation laws, just like the iteration formula for acids([H⁺]) was derived. Still, this conversation helped my thinking, so thanks for that.

 

[epenguin]

I'm getting pH 7.2 as near as makes no difference., with a square rooting but no polynomial equations

[OH^-] - [H⁺] = [Na⁺] - [NO₂^-] = [HNO₂] = [H⁺][NO₂^-] /K_a

I then make the approximation [NO₂^-] = [Na⁺] which, since we have salt of strong base and weak acid and we must be above pH 7 is good to better than 1/1000, and thence get, as the numbers happen to be easy

[HNO₂] = (3/2) [H^+]]

and thence from first line

(5/2) [H⁺] = [OH^-]

and thence via the K_w equation [H⁺] = 0.4×10^-7 and square-rooting etc. finally pH = 7.1989 insignificantly different from 7.2.

I hope you can see these are fairly natural steps.