Calculate the volume of 0.0500 M EDTA needed to titrate 29.13 mL of 0.0598 M Mg(NO3)2?
回答 (1)
2016-02-22 9:42 am
1 mole of Mg^2+ ions reacts with 1 moles of EDTA.
No. of moles of Mg^2+ ions = (0.0598 mol/L) × (29.13/1000 L) = 0.001742 mol
No. of moles of EDTA = 0.001742 mol
Volume of EDTA = (0.001742 mol) / (0.0500 mol/L) = 0.03484 L = 34.84 mL
No. of moles of Mg^2+ ions = (0.0598 mol/L) × (29.13/1000 L) = 0.001742 mol
No. of moles of EDTA = 0.001742 mol
Volume of EDTA = (0.001742 mol) / (0.0500 mol/L) = 0.03484 L = 34.84 mL
收錄日期: 2021-04-18 14:31:29
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