# How can you increase the pH of an acidic solution by 1?

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?

• You can't tell without more data. Mar 1, 2021 at 22:40

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.

• I assume the task author goes for the simplified formula, based on $c_0 \gg [\ce{H+}] \gg [\ce{OH-}]$ Mar 2, 2021 at 2:18
• Hmm yeah I have now added that in a block just now Mar 2, 2021 at 3:23
• why does 𝑥2=10^−(1−𝑙𝑜𝑔([𝑥])) and not x2=10^-log[x]+1
– user105067
Mar 4, 2021 at 20:08
• 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 Mar 5, 2021 at 0:26