# How do you solve titration problems?

Feb 6, 2017

Try using dimensions when you do the calculations.

#### Explanation:

With a titration a measured quantity of titrant is added to a known mass of known molar quantity.

We use the relationship, $\text{moles "="Mass"/"molar mass}$,

and $\text{Molarity (concentration)}$ $=$ $\text{Moles of stuff"/"Volume of solution}$

Now when we use $\text{molarity}$ we can preserve the dimensions: $m o l \cdot {L}^{-} 1$ are the units for concentration.

When we multiply a concentration by a VOLUME, we get the product, $m o l \cdot \cancel{{L}^{-} 1} \times \cancel{L} = m o l$.

Now a chemical reaction is VITAL when you do a titration. For acid/base titrations this is typically:

${H}_{3} {O}^{+} + H {O}^{-} \rightarrow 2 {H}_{2} O \left(l\right)$

The stoichiometric equivalence must be borne in mind, when you do the calculation. So, as a general rule, you must have the appropriate, stoichiometrically balanced chemical equation. Most of the time, there is 1:1 molar equivalence, and,

${C}_{1} {V}_{1} = {C}_{2} {V}_{2}$, where $C \equiv \text{concentration}$ as we defined.