10 g of (molar mass = 100 g/mol) is dissolved in 250 mL of 1 M . After the reaction is complete, the — Redox Reactions and Volumetric Analysis Chemistry Question
Question
10 g of $CaCO_3$ (molar mass = 100 g/mol) is dissolved in 250 mL of 1 M $HCl$. After the reaction is complete, the solution is boiled to expel $CO_2$. What volume (in mL) of 2 M $KOH$ solution would be required to exactly neutralize the excess $HCl$ remaining in the solution?
Answer: 25
💡 Solution & Explanation
Moles of $CaCO_3 = 10 / 100 = 0.1$ mol. The n-factor for $CaCO_3$ neutralizing an acid is 2. Equivalents of $CaCO_3 = 0.1 \times 2 = 0.2$ eq = 200 meq. Total milli-equivalents of $HCl$ initially present = $250 \times 1 = 250$ meq. Milli-equivalents of excess $HCl$ remaining = $250 - 200 = 50$ meq. To neutralize this with $KOH$ (n-factor = 1, so N = 2N): $V_{KOH} \times 2 = 50 \Rightarrow V_{KOH} = 25$ mL.
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