Titration situation is a common context for problems.
Tracking solutions/dilutions/samples
"6g of $Na_{2}CO_{3}$ is made up to 500$mL$ solution. 20$mL$ of this solution requires 21.2$mL$ of $HCl$ to reach end point. What is $c(HCl)$?"
$$\begin{align} n = \frac{m}{M}, \therefore n(Na_{2}CO_{3}) &= \frac{6}{2\times 22.99 + 12.01 + 3\times 16} = 5.66\times 10^{-2}mol \ (3SF)\\ c = \frac{n}{V}, \ \therefore [Na_{2}CO_{3}] &= \frac{5.66\times 10^{-2}}{0.500}=1.13 \times 10^{-1} molL^{-1} \ (3SF) \\ n = cV, \therefore n (\text{reacting} Na_{2}CO_{3}) &= 1.13\times 10^{-1} \times 0.020 = 2.26 \times 10^{-3}mol \ (3SF) \\ Na_{2}CO_{3} + 2HCl &\to 2Na^{+} + CO_{2} + H_{2}O + 2Cl^{-} \\ \therefore \text{1:1 ratio, }n(HCl) &= 2\times n(Na_{2}CO_{3}) = 4.53\times 10^{-3} mol \ (3SF) \\ c = \frac{n}{V}, \therefore [HCl] &= \frac{4.53\times 10^{-3}}{0.0212}=0.2 molL^{-1} \ (1SF) \end{align} $$
