A 672 mg sample of impure sodium carbonate was titrated with 41.06 mL of 0.243 M

Question

A 672 mg sample of impure sodium carbonate was titrated with 41.06 mL of 0.243 M hydrochloric acid. Calculate the percent mass of sodium carbonate in the sample.

Explanation / Answer

find total moles of acid 0.05000 L of 0.200 mol / litre HCl = 0.0100 total moles of acid find moles of NaOH 0.04750 L of 0.125 mol / litre NaOH = 0.005937 moles of NaOH by the equation 1 NaOH & 1 HCl --> 1 NaCl & 1 H2O 0.005937 moles of NaOH reacts with an equal number of moles of HCl = 0.005937 moles of HCl find the moles of HCl that reacted with the NaHCO3 0.0100 total moles of acid - 0.005937 moles of HCl = 0.0040625 moles of HCl reacted with NaHCO3 by the equation 1 HCl & 1 NaHCO3 --> 1 NaCl & 1 H2O & 1 CO2 0.0040625 moles of HCl reacts with an equal number of mol NaHCO3 = 0.0040625 mol NaHCO3 use molar mass to find grams of NaHCO3 0.0040625 mol NaHCO3 @ 84.01 g/mol = 0.3413 grams of NaHCO3 find % in sample 0.3413 grams of NaHCO3 / 0.500 grams of sample = 68.26% your answer is rounded to 3 sig figs 68.3 % NaHCO3

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