Back titration
Insoluble base, e.g. $CaCO_{3}$ which is impure
React with acid - known concentration and volume
Once this reaction is complete, we do another titration to find the concentration of the excess acid.
Suppose 4g of impure $CaCO_{3}$ placed into 100mL of $0.5molL^{-1} \ HCl$. We titrate the excess acid with 20 mL of $1.00molL^{-2}$ $NaOH$.
$$\begin{align} n(HCl) = cV = 0.5\times 0.1 = 0.05mol\\ n(NaOH)= cV = 1 \times 0.02 = 0.02mol \\ 1:1 ratio, \therefore n(HCl ) \text{ reacted} = n(NaOH) =0.02mol\\ \therefore n(HCl)\text{ consumed} = 0.05 - 0.02 = 0.03mol \\ 1:2 ratio, \therefore n(CaCO_{3} ) = \frac{1}{2} n(HCl) \text{consumed} = 0.015mol \\ \therefore m(CaCO_{3} ) = nM = 0.015 \times (40.08+12.01+3\times 16) = 1.50g \\ \therefore \text{\\% purity} = \frac{1.5}{4}\times 100 = 37.5\\% (3SF) \end{align} $$
