5.3 g of a metal carbonate is dissolved in 150 mL of 1 N . The unused acid required 100 mL of 0.5 N — Redox Reactions and Volumetric Analysis Chemistry Question

💡 Solution & Explanation

Total milli-equivalents of $HCl$ = $150 \times 1 = 150$ meq. Milli-equivalents of $HCl$ neutralized by $NaOH$ = $100 \times 0.5 = 50$ meq. Therefore, meq of $HCl$ reacted with $M_2CO_3$ = $150 - 50 = 100$ meq = 0.1 eq. Thus, 0.1 equivalents of $M_2CO_3$ weigh 5.3 g. Equivalent weight of the salt $M_2CO_3$ = $5.3 / 0.1 = 53$. Since $E_{salt} = E_{metal} + E_{carbonate}$, $53 = E_M + (60/2) \Rightarrow E_M = 53 - 30 = 23$.