40mathrm~mL$40\mathrm{~mL}$ of a mixture of mathrmCH_3mathrmCOOH$\mathrm{CH}_{3}\mathrm{COOH}$ and mathrmHCl$\mathrm{HCl}$ (aqueous solution) is titrated against 0.1mathrm~M~NaOH$0.1\mathrm{~M~NaOH}$ solution conductometrically. Which of the following statements is correct?

### Related Formula At the equivalence point during titration: M_textacid V_textacid = M_textbase V_textbase$$M_{\text{acid}} V_{\text{acid}} = M_{\text{base}} V_{\text{base}}$$ ### Core Logic In a mixture of a strong acid (mathrmHCl$\mathrm{HCl}$) and a weak acid (mathrmCH_3mathrmCOOH$\mathrm{CH}_3\mathrm{COOH}$): 1. mathrmHCl$\mathrm{HCl}$ is a strong acid and is completely ionized. When mathrmNaOH$\mathrm{NaOH}$ is added, highly mobile mathrmH^+$\mathrm{H}^+$ ions are replaced by less mobile mathrmNa^+$\mathrm{Na}^+$ ions, causing a sharp drop in conductance (segment AB). 2. At point B (2.0mathrm~mL$2.0\mathrm{~mL}$), mathrmHCl$\mathrm{HCl}$ is completely neutralized. 3. Segment BC represents the neutralization of the weak acid mathrmCH_3mathrmCOOH$\mathrm{CH}_3\mathrm{COOH}$ to form highly conducting sodium acetate, causing a moderate rise in conductance up to point C (5.0mathrm~mL$5.0\mathrm{~mL}$). 4. Beyond point C, excess mathrmOH^-$\mathrm{OH}^-$ ions cause a rapid rise in conductance (segment CD). ### Step 1: Calculate concentration of mathrmHCl$\mathrm{HCl}$ Volume of mathrmNaOH$\mathrm{NaOH}$ used to neutralize mathrmHCl$\mathrm{HCl}$ is V_1 = 2.0mathrm~mL$V_1 = 2.0\mathrm{~mL}$: M_mathrmHCl times 40mathrm~mL = 0.1mathrm~M times 2.0mathrm~mL$$M_{\mathrm{HCl}} \times 40\mathrm{~mL} = 0.1\mathrm{~M} \times 2.0\mathrm{~mL}$$ M_mathrmHCl = frac0.240 = 0.005mathrm~M$$M_{\mathrm{HCl}} = \frac{0.2}{40} = 0.005\mathrm{~M}$$ ### Step 2: Calculate concentration of mathrmCH_3mathrmCOOH$\mathrm{CH}_3\mathrm{COOH}$ Volume of mathrmNaOH$\mathrm{NaOH}$ used to neutralize mathrmCH_3mathrmCOOH$\mathrm{CH}_3\mathrm{COOH}$ is V_2 = 5.0mathrm~mL - 2.0mathrm~mL = 3.0mathrm~mL$V_2 = 5.0\mathrm{~mL} - 2.0\mathrm{~mL} = 3.0\mathrm{~mL}$: M_mathrmCH_3mathrmCOOH times 40mathrm~mL = 0.1mathrm~M times 3.0mathrm~mL$$M_{\mathrm{CH}_3\mathrm{COOH}} \times 40\mathrm{~mL} = 0.1\mathrm{~M} \times 3.0\mathrm{~mL}$$ M_mathrmCH_3mathrmCOOH = frac0.340 = 0.0075mathrm~M$$M_{\mathrm{CH}_3\mathrm{COOH}} = \frac{0.3}{40} = 0.0075\mathrm{~M}$$ ### Pattern Recognition Conductometric titration curves are analyzed sequentially: the strongest electrolyte is always neutralized first. A steep drop in conductance always signals the neutralization of a strong acid (mathrmH^+$\mathrm{H}^+$ depletion). A weak acid titration shows a gentle upward slope due to salt formation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry Class 11 Chemistry: Equilibrium