100 mL of 0.2 M solution is mixed with 400 mL of 0.1 M solution. What is the final total molar conce — Redox Reactions and Volumetric Analysis Chemistry Question
Question
100 mL of 0.2 M $AlCl_3$ solution is mixed with 400 mL of 0.1 M $HCl$ solution. What is the final total molar concentration of all cations and chloride ions, respectively?
Answer: A
💡 Solution & Explanation
Cations: $Al^{3+} = 100 \times 0.2 = 20$ mmol; $H^+ = 400 \times 0.1 = 40$ mmol. Total cations = 60 mmol. Total volume = 500 mL. Cation concentration = $60 / 500 = 0.12 M$. Anions: $Cl^-$ from $AlCl_3 = 3 \times 20 = 60$ mmol; $Cl^-$ from $HCl = 40$ mmol. Total $Cl^- = 100$ mmol. Chloride concentration = $100 / 500 = 0.2 M$.
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