25 mL of the given HCl solution requires 30 mL of 0.1M sodium carbonate solution. What is the volume of this HCl solution required to titrate 30 mL of 0.2M aqueous NaOH solution?

First we find the strength of the unknown HCl solution by using the information given for its titration with sodium carbonate.

The balanced chemical equation for the reaction between hydrochloric acid and sodium carbonate is

$$2\text{HCl}+ \text{Na}_2\text{CO}_3 \;\longrightarrow\; 2\text{NaCl}+ \text{H}_2\text{O}+ \text{CO}_2$$

From the equation we see that 2 moles of HCl react with 1 mole of Na 2CO 3.

Now we calculate the moles of Na 2CO 3 actually used:

$$\text{Molarity (M)} = \dfrac{\text{moles}}{\text{volume in litres}}$$

Given molarity of Na 2CO 3 is $$0.1\,\text{M}$$ and its volume is $$30\,\text{mL}=30\times10^{-3}\,\text{L}$$, so

$$\text{moles of Na}_2\text{CO}_3 = 0.1 \times 30\times10^{-3} = 0.003\ \text{mol}$$

Because $$2$$ moles of HCl are needed per mole of Na 2CO 3, the moles of HCl that reacted are

$$\text{moles of HCl} = 2 \times 0.003 = 0.006\ \text{mol}$$

These 0.006 mol of HCl are present in the 25 mL (that is, $$25\times10^{-3}\,\text{L}$$) portion that was taken. Hence the molarity of the HCl solution is

$$M_{\text{HCl}} = \dfrac{0.006}{25\times10^{-3}} = \dfrac{0.006}{0.025} = 0.24\ \text{M}$$

Next we use this 0.24 M HCl solution to titrate an NaOH solution.

The neutralisation equation is

$$\text{HCl}+ \text{NaOH} \;\longrightarrow\; \text{NaCl}+ \text{H}_2\text{O}$$

Here the stoichiometric ratio is 1:1; one mole of HCl neutralises one mole of NaOH.

Calculate the moles of NaOH present in the 30 mL, 0.2 M solution:

$$\text{moles of NaOH} = 0.2 \times 30\times10^{-3} = 0.006\ \text{mol}$$

Therefore we need 0.006 mol of HCl. Using the molarity just found, the required volume is

$$V_{\text{HCl}} = \dfrac{\text{moles}}{\text{molarity}} = \dfrac{0.006}{0.24}\ \text{L}$$

$$V_{\text{HCl}} = 0.025\ \text{L} = 25\ \text{mL}$$

Hence, the correct answer is Option A.