Calculating pH for monoprotic acids

This note could have been named “everything you need to know to calculate pH for any acid in this course” (or for any base).

pH of an acid

Strong Monoprotic Acid

If you have a $c_{0}$ molar solution of a strong acid $H A$ , you can assume that proton transfer with water occurs completely:

HA + H X_{2} O A X^{-} + H X_{3} O X^{+} (1)

As a result, $[A X^{-}]_{eq} = [H X_{3} O X^{+}]_{eq} = c_{0}$ , so $p H = - lo g [H X_{3} O X^{+}]_{eq} = - lo g c_{0}$ . Remarkably simple. Don’t forget that the log is base 10. I guess the only thing you may worry about is whether the log on your calculator is base 10 or base $e$ , but it’s easy to check: calculate $lo g 10$ and you should get a 0. If you don’t get a 0, try calculating $lo g 2.71$ - if that’s closer to zero, your log is natural log. Just as a reminder, logs satisfy the following property:

lo g_{a} b = \frac{lo g _{c} b}{lo g _{c} a}

so if we’re interested in log base 10 (i.e. $a = 10$ ) and we only have natural log (i.e. $c = e$ ), then:

lo g b = \frac{ln b}{ln 10}

Weak Monoprotic Acid

If you have $c_{0}$ molar solution of a weak acid $H A$ , proton transfer occurs partially:

HA + H X_{2} O A X^{-} + H X_{3} O X^{+} (1)

There’s a standard algorithm to solve problems like this. We assume that everything occurs in a solution of volume $V_{i}$ . In this solution, initially there are $c_{0} V_{i}$ moles of $HA$ and $0$ moles of both $A X^{-}$ and $H X_{3} O X^{+}$ . Now, let’s suppose that $x$ moles of $HA$ transfers proton to $H X_{2} O$ . As a result, by stoichiometry, $x$ moles of $H X_{3} O X^{+}$ and $A X^{-}$ form. Let’s make a table:

	$HA$	$A X^{-}$	$H X_{3} O X^{+}$
Initial num. moles	$c_{0} V_{i}$	0	0
Reacted	$- x$	$+ x$	$+ x$
Final num. moles	$c_{0} V_{i} - x$	$x$	$x$

Whenever you’re asked to calculate pH of a solution, it’s implied that the proton transfer reaction reaches equilibrium. That is, the ratio of concentrations should be equal to $K_{a}$ :

K_{a} = \frac{[ A X ^{-} ] [ H X _{3} O X ^{+} ]}{[ HA ]} (2)

To calculate equilibrium concentrations of all molecules, we need to divide the number of moles by concentration. Let’s take the final volume as $V_{f}$ :

[HA] = \frac{c _{0} V _{i} - x}{V _{f}} [H X_{3} O X^{+}] = \frac{x}{V _{f}} [A X^{-}] = \frac{x}{V _{f}}

A common (and fairly valid assumption) is that the volume of the solution doesn’t change in the course of proton transfer. So $V_{f} = V_{i}$ . Then:

[HA] = \frac{c _{0} V _{i} - x}{V _{i}} = c_{0} - \frac{x}{V _{i}} [H X_{3} O X^{+}] = \frac{x}{V _{i}} [A X^{-}] = \frac{x}{V _{i}}

You might notice that we can rewrite $[HA]$ as:

[HA] = c_{0} - \frac{x}{V _{i}} = c_{0} - [H X_{3} O X^{+}]

As a result, the expression for $K_{a}$ becomes:

K_{a} = \frac{[ H X _{3} O X ^{+} ] ^{2}}{c _{0} - [ H X _{3} O X ^{+} ]} (3)

because $c_{0}$ and $K_{a}$ are known, we can solve for $[H X_{3} O X^{+}]$ and then find the pH.

A small elephant

The solution with explicit consideration of the volume above may seem a bit unorthodox to you; after all, why can’t we just say let’s say we had $c_{0}$ M of $HA$ , then $x$ M reacted:

	$HA$	$A X^{-}$	$H X_{3} O X^{+}$
Initial concentration	$c_{0}$	0	0
Reacted	$- x$	$+ x$	$+ x$
Final concentration	$c_{0} - x$	$x$	$x$

and so $K_{a}$ becomes:

K_{a} = \frac{x ^{2}}{c _{0} - x} (4)

seeing that $[H X_{3} O X^{+}] = x$ , you know that if you solve for $x$ , you can find $[H X_{3} O X^{+}]$ .

The problem with this approach is that it’s fundamentally incorrect. You cannot add (or subtract) concentrations. Concentrations (as most other relative properties) are not additive. You can add/subtract moles simply because moles is a direct metric of countable quantity. 5 apples plus 3 apples is 8 apples. 5 molecules plus 3 molecules is 8 molecules.

Now tell me how many apples per bag will you have if I give you 5 apples per bag and 3 apples per bag? Can you answer that question without me specifying how many bags do I give you? No. What if I give you 2 bags with 10 apples (5 apples per bag) and another bag with 3 apples? Or what if I give you 1 bag with 5 apples and 1 bag with 3 apples. The answer inherently depends on the number of bags and in either case you’ll not get 8 apples per bag. The only way to get 5 apples/bag + 3 apples/bag to be equal to 8 apples/bag is if you keep only one bag and add every apple to it.

Translating that into concentrations, the only case when you can add/subtract concentrations is if the volume doesn’t change. Which is true, in this particular case. And I guess the conventional approach is to ignore such nuances and simply tell you (the students) to subtract concentrations, just to keep things simple. You lose that simplicity whenever the volume starts to change (as in when dealing with buffers), and the conventional approach is to say, now you have to consider volume. I personally think it’s better to do everything properly from the beginning, because then, you can confidently say that every calculation in acid-base equilibria is done in the same way (which it is), there’s no conceptual difference between pure acids/bases or buffers. Instead of adding the anxiety of having to decide whether volume should be considered or not (or even forgetting about it and losing points for that). I’m genuinely curious to know which option would you prefer.

Just to be clear

Regardless of whether you do it properly and consider the volume or directly subtract concentrations, the end result is the equation:

K_{a} = \frac{[ H X _{3} O X ^{+} ] ^{2}}{c _{0} - [ H X _{3} O X ^{+} ]} (3)

Which can be rearranged:

c_{0} K_{a} - K_{a} [H X_{3} O X^{+}] [H X_{3} O X^{+}]^{2} + K_{a} [H X_{3} O X^{+}] - c_{0} K_{a} [H X_{3} O X^{+}]_{\pm} = [H X_{3} O X^{+}]^{2} = 0 = \frac{- K _{a} \pm K _{a}^{2} + 4 K _{a} c _{0}}{2}

$[H X_{3} O X^{+}]$ has to be positive, so the only physically meaningful root is $[H X_{3} O X^{+}]_{+}$ . There’s your solution to literally any monoprotic acid.

By the way, remember Dr. Cooper said that sometimes you can approximate $c_{0} - [H X_{3} O X^{+}]$ in the denominator as approximately equal to $c_{0}$ , simplifying calculations quite significantly? If you wonder when such simplification is justified, check out this page which dynamically shows you pH calculated exactly and with approximation $c_{0} - [H X_{3} O X^{+}] \approx c_{0}$ for different c0 and pKa values.

H2SO4

or diprotic acid which is strong on the first stage. Basically, the idea is that if you start with $c_{0}$ of $H X_{2} SO X_{4}$ , all of it will react with water to form $[HSO X_{4} X^{-}] = [H X_{3} O X^{+}] = c_{0}$ . Then you have to treat $HSO X_{4} X^{-}$ as a weak acid $HA$ and so you proceed with typical calculations with one subtle difference:

	$HA$	$A X^{-}$	$H X_{3} O X^{+}$
Initial concentration	$c_{0}$	0	$c_{0}$
Reacted	$- x$	$+ x$	$+ x$
Final concentration	$c_{0} - x$	$x$	$c_{0} + x$

Instead of initial concentration of $H X_{3} O X^{+}$ being zero, it’s equal to $c_{0}$ . As a result, you get a slightly different quadratic equation:

K_{a} = \frac{x ( c _{0} + x )}{c _{0} - x}

Which you can easily solve.

pH of a base

Strong Base

It’s symmetric to the case with an acid. If you have $c_{0}$ of some strong Arrhenius base (I doubt you’ll see a strong Bronsted Lowry base), then you can assume that $[OH X^{-}] = c_{0}$ . From which you can find pOH, and then pH knowing that pH + pOH = 14. Cool.

Weak base

For some base $B$ :

B + H X_{2} O BH X^{+} + OH X^{-}

write the $K_{b}$ :

K_{b} = \frac{[ BH X ^{+} ] [ OH X ^{-} ]}{[ B ]}

then set up the accounting table: before/initial, reacted, and then final.

Chem165 Notes

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