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Help with a hard titration problem (Back-titration)?

A 0.3240g sample of impure Na2CO3 was dissolved in 50.00mL of 0.1280M HCl. The excess acid then requires 30.10mL of 0.1220M NaOH for complete neutralization. Calculate the % Na2CO3 (MM = 105.99) in the sample.

If you guys want, you can just tell me what i need to do step by step and i'll be the one to do the calculations so i learn.

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moles HCl = 0.05000 L x 0.1280 M= 0.006400

moles NaOH = 0.03010 L x 0.1220 M=0.003672

moles HCl used = 0.006400 - 0.003672=0.002728

Na2CO3 + 2 HCl = H2O + CO2 + 2 NaCl

moles Na2CO3 = 0.002728/2=0.001364

mass Na2CO3 = 0.001364 mol x 105.9876 g/mol=0.1446 g

% = 0.1446 x 100/ 0.3240=44.63