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In order to increase the pH of a solution of some acid $\ce{HA}$ by 1, how many times must it be diluted by? The pH log scale suggests that the required dilution factor is 10. Is that generally valid for strong and/or weak acids, or do we need to take into account the dissociation constant for weak acids?

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    $\begingroup$ You can't tell without more data. $\endgroup$ Mar 1, 2021 at 22:40

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Since you have not given any values some of the important values need to be considered as variables,

Also assuming its a monobasic acid $\ce{HA}$ $$\ce{HA_{aq} <=> H^+_{aq} + A^-_{aq}}$$

The important values are the

  • Acid dissociation constent $K_\mathrm{a}$
  • Initial concentration $c$

Writing the equation for the acid dissociation constant $$K_\mathrm{a} = \ce{\frac{[H^{+}][A^{-}]}{[HA]}}$$

Initially a $x$ amount of acid gets dissociated equation simplifies to $$K_\mathrm{a} = \frac{x^{2}}{(c-x)}$$

This is a $ax^2+bx+c=0$ type equation which can be solved easily simplified, $$x^2 + K_\mathrm{a}x - K_\mathrm{a}c = 0$$

Take the accepted value for $x$ then (Not thinking that you will get complex number for real $K_\mathrm{a}$ and $c$ values),

Optionally you can consider that $c \gg x$ so $c - x \approx c$ simplified $x = \sqrt{K_\mathrm{a}c}$

Since $x$ is found now, we now know the initial $\mathrm{pH}$ value or $\mathrm{pH}_1$ $$\mathrm{pH}=-\log(\ce{[H^+]})$$

so, $$\mathrm{pH}_1=-\log([x])$$

Final $\mathrm{pH}$ value $\mathrm{pH}_2$ goes up by 1 so, $$\mathrm{pH}_2=-\log([x]) + 1$$

So the new dissociated $x_2$ value, $$x_2 = 10^{-(1-\log([x]))}$$

Since $\ce{[H^+]=[A^-]}$ new equation, $$K_\mathrm{a} = \frac{x_{2}^{2}}{(c_{2}-x_{2})} = \frac{10^{-2(1-\log(x))}}{(c_{2}-10^{-(1-\log(x))})}$$

Now you can solve this equation and get $c_2$ new concentration now the answer is almost complete just need to get the dilution factor

$D = \dfrac{c}{c_{2}}$

$D$ is how many times you need to dilute

But in the non monobasic acids this is a bit more complex since it has more than one dissociation constant $K_{\mathrm{a}_{1}}$ $K_{\mathrm{a}_{2}}$ $K_{\mathrm{a}_{3}}$ etc. need to check if the ratios of the dissociation constants if so that they are effective in the $\mathrm{pH}$ calculation.

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    $\begingroup$ I assume the task author goes for the simplified formula, based on $c_0 \gg [\ce{H+}] \gg [\ce{OH-}]$ $\endgroup$
    – Poutnik
    Mar 2, 2021 at 2:18
  • $\begingroup$ Hmm yeah I have now added that in a block just now $\endgroup$
    – Avon97
    Mar 2, 2021 at 3:23
  • $\begingroup$ why does ๐‘ฅ2=10^โˆ’(1โˆ’๐‘™๐‘œ๐‘”([๐‘ฅ])) and not x2=10^-log[x]+1 $\endgroup$
    – user105067
    Mar 4, 2021 at 20:08
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    $\begingroup$ Its just $\ce{pH = -\log{([H^{+}])}}$ substitute the new $\ce{pH_2}$ to the equation and try to get the new $\ce{H^+}$ concentration you made a mathematical mistake watch out there is a minus sign there $\endgroup$
    – Avon97
    Mar 5, 2021 at 0:26

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