Describe how a sample of Iron (II) Carbonate can be prepared using dilute hydrochloric acid.

$$n = c \times V$$

where $n$ is number of moles (mol), $c$ is concentration (mol dm$^{-3}$), $V$ is volume (dm$^3$).

$$n(\ce{HCl}) = 0.1\ \textrm{mol dm}^{-3} \times 0.025\ \textrm{dm}^3 = 0.0025\ \textrm{mol}$$

0.0025 mol

Question 4(b): Number of moles of NaOH that reacted with HCl

$$n(\ce{HCl, back}) = 0.1\ \textrm{mol dm}^{-3} \times \frac{24.9}{1000}\ \textrm{dm}^3$$ $$n(\ce{HCl, back}) = 0.1 \times 0.0249 = 0.00249\ \textrm{mol}$$

Excess $\ce{NaOH} = 0.00249\ \textrm{mol}$.

$$n(\ce{NaOH, total}) = n(\ce{HCl, pipetted}) + n(\ce{HCl, back})$$ $$n(\ce{NaOH, total}) = 0.0025\ \textrm{mol} + 0.00249\ \textrm{mol} = 0.00499\ \textrm{mol}$$

$$n(\ce{NaOH, reacted}) = n(\ce{NaOH, total}) - \textrm{excess NaOH}$$ $$n(\ce{NaOH, reacted}) = 0.00499\ \textrm{mol} - 0.00249\ \textrm{mol} = 0.0025\ \textrm{mol}$$

0.0025 mol

Question 4(c): Concentration of NaOH in moles per dm$^3$

$$c(\ce{NaOH}) = \frac{0.00499\ \textrm{mol}}{0.0250\ \textrm{dm}^3} = 0.20\ \textrm{mol dm}^{-3}$$

0.20 mol dm^{-3}