Titration - how the math works.

As explained here, titration allows us to find the concentration of a solution by reacting it with another solution, about which we have more information. 

Most commonly, titration is done with acids and bases, so we’ll discuss those first and then discuss other possibilities.

An acid base reaction is “done” when the number of hydroxide ions matches the number of hydronium ions. In other words, the reaction is done when moles of \(H^{+1}\) = moles of \(OH^{-1}\). Assuming that the acid and base are both monoprotic (that is the acid only has one H+ ion to donate and the base can only take one H+ ion) then we can say that the reaction is “done” when

\(mole _{acid} = mole _{base}\)

If we are working with solutions, then \(moles = concentration ( \frac{mol}{L})\cdot Volume(L)\). So,

\(mole_a = mole_b ~~becomes~~ M_aV_a = M_bV_b\)

In all cases, titration is done so that three of these values are known and the fourth can, therefore be found. Here’s an example:

25.00 mL of an acid with an unknown concentration are titrated with 32.64 mL of 0.445 M NaOH. What is the concentration of the acid?

In this case we know everything except the \(M_a\) so we can solve for that algebraically, getting

\(M_a = \frac{M_bV_b}{V_a}\)

substituting in our values, we get 

\(\frac{(0.445M)\cdot(32.64 mL)}{(25.00mL)}\) = 0.581 M

Note that although this math should be done with liters, we can “get away with” using mL since the two units will cancel each other out.

Dealing with Solids 

It is important to remember that the equation that matters is NOT \(M_aV_a= M_bV_b\), but rather, \(moles_a = moles_b\), That means that we can use other ways to get moles. If, for instance, the acid is a solid, then \(moles_a = \frac{mass_a}{Molar~Mass_a}\). If the base was still a solution, that would give us the following: 

\(\frac{mass_a}{MM_a} = M_bV_b\)

Of course the base could be the solid. The point is not to "learn" the equation above, but rather to understand that the "moles" on either side of the equation can be ANYTHING that translates into moles.

Dealing with diprotic and triprotic acids and bases 

To understand the impact that di- and triprotic acids and bases have on the math of titrations, you need to remember that the equation is NOT moles acid = moles base, but rather moles \(H^{+1}\) ions = moles \(OH^{-1}\) ions.

Imagine that you had 30.0 mL of 0.500 M \(H_3PO_4\). If you needed the moles of phosphoric acid, you could calculate \(M_aV_a = (0.500\frac{mol}{L}) \cdot (0.0300 L) = 0.015 mols H_3PO_4\). However, in a titration, the phosphoric acid has 3 hydrogens that can react with the base, so the moles of \(H^{+1}\) ions is 3 times the moles of \(H_3PO_4\). 

The same idea plays out with a polyprotic base. For instance, 0.12 moles of \(Mg(OH)_2\) can produce 0.240 moles of hydroxide ions. We can compensate for this by adding in a factor on both sides of the equation, so \(M_aV_a= M_bV_b\) is more correctly written as \((\#_a)M_aV_a = (\#_b)M_bV_b\) where \(\#_a\) is the number of H’s in the acid and \(\#_b\) is the number of OH’s in the base.

Here's a sample problem involving polyprotic acids and bases:

How many mL of a 0.115 M solution of barium hydroxide are needed to fully titrate 18.5 mL of 0.331 M phosphoric acid?

Barium hydroxide is \(Ba(OH)_2\) meaning that each mole will produce 2 moles of \(OH^{-1}\) ions. Phosphoric acid is \(H_3PO_4\), so each mole of phosphoric acid will result in 3 moles of \(H^{+1}\) ions. Our math, therefore, looks like this:

\((\#_a)M_aV_a = (\#_b)M_bV_b\)

We are trying to solve for the volume of base needed, so we can rearrange algebraically which will give us:

\(V_b=\frac{(\#_a)M_aV_a}{(\#_b)M_b}\)

Substituting in our values from the problem, gives us this:

\(V_b=\frac{(3)(0.331M)(18.5 mL)}{(2)(0.115M)}=79.8 mL\)