- #1
TheSodesa
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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?
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