10mathrm~mL$10\mathrm{~mL}$ of 2mathrm~M~NaOH$2\mathrm{~M~NaOH}$ solution is added to 20mathrm~mL$20\mathrm{~mL}$ of 1mathrm~M~HCl$1\mathrm{~M~HCl}$ solution kept in a beaker. Now, 10mathrm~mL$10\mathrm{~mL}$ of this mixture is poured into a volumetric flask of 100mathrm~mL$100\mathrm{~mL}$ containing 2 moles of mathrmHCl$\mathrm{HCl}$ and made the volume upto the mark with distilled water. The solution in this flask is :

### Related Formula Number of millimoles (n$n$) is given by: n = M times V_mathrmmL$$n = M \times V_{\mathrm{mL}}$$ Molarity (M$M$) of a diluted mixture: M = fractextTotal molestextTotal Volume in Liters$$M = \frac{\text{Total moles}}{\text{Total Volume in Liters}}$$ ### Core Logic Evaluate the first mixing step to determine the net acid-base state: - Millimoles of mathrmNaOH = 10mathrm~mL times 2mathrm~M = 20mathrm~mmol$\mathrm{NaOH} = 10\mathrm{~mL} \times 2\mathrm{~M} = 20\mathrm{~mmol}$ - Millimoles of mathrmHCl = 20mathrm~mL times 1mathrm~M = 20mathrm~mmol$\mathrm{HCl} = 20\mathrm{~mL} \times 1\mathrm{~M} = 20\mathrm{~mmol}$ Since millimoles are equal, mathrmHCl$\mathrm{HCl}$ and mathrmNaOH$\mathrm{NaOH}$ completely neutralize each other, producing a neutral aqueous salt solution. ### Step 1: Analyze transfer to volumetric flask Taking 10mathrm~mL$10\mathrm{~mL}$ of this neutralized solution provides no excess mathrmH^+$\mathrm{H}^+$ or mathrmOH^-$\mathrm{OH}^-$ ions. It is added to a volumetric flask containing 2mathrm~mol$2\mathrm{~mol}$ of pure mathrmHCl$\mathrm{HCl}$. ### Step 2: Calculate final molarity of mathrmHCl$\mathrm{HCl}$ The volume of the flask is made up to 100mathrm~mL = 0.1mathrm~L$100\mathrm{~mL} = 0.1\mathrm{~L}$: M = frac2mathrm~mol0.1mathrm~L = 20mathrm~M$$M = \frac{2\mathrm{~mol}}{0.1\mathrm{~L}} = 20\mathrm{~M}$$ Hence, the resulting solution is 20mathrm~M~HCl$20\mathrm{~M~HCl}$. ### Pattern Recognition Stoichiometric neutralizations are evaluated by setting up mole/millimole balance charts. Once stoichiometric equivalence (MV_textacid = MV_textbase$MV_{\text{acid}} = MV_{\text{base}}$) is reached, any sub-aliquot of that solution remains completely neutral. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry Class 11 Chemistry: Ionic Equilibrium