0.4 grams sample of primary standard Na2CO3 treated with 45mL of dilute perchlor

Question

0.4 grams sample of primary standard Na2CO3 treated with 45mL of dilute perchloric acid. Solution boiled to remove CO2 and then the excess HClO4 back titrated with 13mL of dilute NaOH. In a seperate experiment 18.44 mL of HClO4 neutralized NaOH in a 16.0 mL portion. Calculate molarities of HClO4 and NaOH

Explanation / Answer

0.4 g Na2CO3 @ 105.99 g/mol = 0.003774 moles Na2CO3 ================ 1 Na2CO3 & 2 HClO3 --> 0.003774 mol Na2CO3 @ 2 mol HClO3 / 1 mol NaOH = 0.007548 mol HClO3 reacts with the Na2CO3 ===================== 18.44 ml HClO3 reacts with 16 ml of NaOH therefor in the back titration : 13 ml NaOH @ 18.44 ml HClO3 / 16 ml NaOH = 14.98 ml of HClO3 needed back titrating therfor in the reaction with Na2CO3, 45 ml - 14.98 ml = 30.02 ml of HClO3 reacted with the Na2CO3 therefor the 0.007548 mol HClO3 are in the 30.02 ml of HCl O3: 0.007548 mol HClO3 / 0.03002 L = 0.2514 Molar your first answer: 0.2514 Molar HClO3 ======================================… 18.44 ml HClO3 reacts with 16 ml of NaOH 0.01844 L HClO3 @ 0.2514 mol/Litre HClO3 = 0.004636 moles HClO3 0.004636 moles HClO3 = 0.004636 moles NaOH (in the 16 ml) 0.004636 moles NaOH / 0.016 litres NaOH = 0.2897 Molar second answer: NaOH is 0.2897 Molar ============= you need to review the data provided & to determine how many significant digits this experiment has; & then round off your answers. "18.44" had 4sigfigs 45, 13, & 16 only have 2 sig figs but 0.4 g Na2CO3 has only 1 sig fig

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