25\,mL of the given HCl solution requires 30\,mL of 0.1\,M sodium carbonate solution. What is the volume of this HCl…

🧠 Step 1: find HCl strength. Step 2: titrate the base. First, $\ce{Na2CO3 + 2HCl -> 2NaCl + H2O + CO2}$ gives $1$ carbonate : $2$ acid. Use this to fix the molarity of HCl. Then use $\ce{NaOH + HCl -> NaCl + H2O}$ to find the volume needed.

🗺️ Two-step titration Step 1: moles of $\ce{Na2CO3} = 30 \times 0.1 = 3\,\text{mmol}$. Acid needed $= 2 \times 3 = 6\,\text{mmol}$ in $25\,\text{mL}$. $M_{\ce{HCl}} = \frac{6}{25} = 0.24\,\text{M}.$ Step 2: moles of NaOH $= 30 \times 0.2 = 6\,\text{mmol}$. Acid needed = $6\,\text{mmol}$. $V = \frac{6}{0.24} = 25\,\text{mL}.$

⚡ Speed scan $0.1\,\text{M}$ $\ce{Na2CO3}$ neutralises like $0.2\,\text{M}$ $\ce{NaOH}$ acid-wise. Same volume of base ($30\,\text{mL}$), so same volume of HCl: $25\,\text{mL}$.

$\boxed{\text{Answer: (a)}}$