Doubt on process to calculate pKa [closed]

Question

I have to solve an exercise that I've been trying for a long time without success.

Information that I have:

• pH @ certain temperature of an aqueous dissolution of an acid
• molar concentration of the acid

Unknown:

• approximate $\mathrm{p}K_\mathrm{a}$ (is not meant to be calculated with calculator I suppose that is the reason they say "approximate")

So far I tried to express the chemical equation and its two steps to finally get $\ce{H3O+}$ and an oxid. I've calculated backwards the Hydronium concentration $[\ce{H3O+}]$ with the $\mathrm{pH}$ formula. $$ \mathrm{pH} = -\log([\ce{H3O+}])$$. I've written down the formula to calculate $K_\mathrm{a}$ from where I expected to be able to calculate $\mathrm{p}K_\mathrm{a}$.

I don't know how to extract now the concentrations of every component on the $K_\mathrm{a}$ formula in order to be able to calculate that. Moreover, what am I suppose to do with the temperature?

I hope you can give me guidance in the process to solve such exercises.

Answer 1

• approximate $\mathrm{p}K_\mathrm{a}$ (is not meant to be calculated with calculator I suppose that is the reason they say "approximate")

The first approximation: the equilibrium concentrations of $\ce{H2SO3}$ and of $\ce{HSO3−}$ are controlled solely by the equilibrium $$\ce{H2SO3 <=> H+ + HSO3−}$$ Because the first $K_\mathrm{a}(K_\mathrm{a1} = 0.017$) is much larger than the second$(K_\mathrm{a2}=1\times{10^{−7.19}})$

The second approximation:$\ce[\ce{H+}] \approx [\ce{HSO3-} ]$

Answer 2

Say your acid is monobasic: suppose its formula is $\ce{HA}$. So, it dissociates according to following equation: $$\ce{HA + H2O <=> H3O+ + A-}$$ Suppose its molar concentration (given) is $c$, and it dissociate $\alpha$ amount in water (assume total volume is $\pu{1.0 L}$) at given temperature at equilibrium. Thus, equilibrium concentrations of $\ce{HA, H3O+}$, and $\ce{A-}$ are $(c-\alpha)$, $\alpha$, and $\alpha$, respectively.

$$\therefore \; K_\mathrm{a} = \frac{[\ce{H3O+}][\ce{A-}]}{[\ce{HA}]}=\frac{\alpha \cdot \alpha}{c-\alpha}$$

Since $\alpha = [\ce{H3O+}]$, $-\log \alpha = \mathrm{pH}$ (which is given) and you know $c$, which is also given, you can estimate the $\mathrm{p}K_\mathrm{a}$ (and hence $K_\mathrm{a}$ if you needed).

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Late edition:

Take (-$\log$) on both side of above equation. Thus:

$$\mathrm{p}K_\mathrm{a} = -2\log \alpha + \log{(c-\alpha)}=2(\mathrm{pH})+ \log{(c-\alpha)}$$ Most of the time (in cases of $c\gt\gt\alpha$), $\mathrm{p}K_\mathrm{a} \approx 2(\mathrm{pH})$

Answer 3

Thank you very much. Now I am gonna insert the values and show you what I get. I have no clue if it makes sense or not.

I start with an aquous dissolution of $\ce{H2SO3}$. I know it will dissociate in two steps, however I read in a book that the first dissociation is normally the strongest and almost every time you can easily ignore the second one, so I proceeded like it follows:

$$p\mathrm{H} = -\log([\ce{H3O+}]) = 1.1$$

This value was given. Then we know that the dissolution has a concentration of $0.5\,\mathrm{M}$. Thus:

$$\ce{H2SO3 + H2O -> H3O+ + HSO3-}$$ $$\ce{c - x -> x +x}$$

$$K_a = \frac{[\ce{H3O+}][\ce{HSO3-}]}{[\ce{H2SO3}]}=\frac{x·x}{c-x}=\frac{x^2}{c-x}$$

I don't know how to determine this without a calculator. Is there an approximation for this?

$$K_a = \frac{10^{-2.2}} {0.5-10^{-1.1}} = 0.015$$

From this, and using the expression by the last equation:

$$pK_a = -2 \log(x) + \log(c-x) = 2.2 + \log(0.5-10^{-1.1}) = 1.82$$

Does that make any sense to you?