What is the weight percent concentration of sodium carbonate in your unknown if

Question

What is the weight percent concentration of sodium carbonate in your unknown if it takes 37.25 mL of 0.1018 M hydrochloric acid to reach the second equivalence point of the titration of a 0.5376 gram sample of the unknown?<?xml:namespace prefix = o ns = "urn:schemas-microsoft-com:office:office" />

Explanation / Answer

Titration equation: CO3{2-} + 2H3O{+} -----> H2CO3 + 2H2O since 2 mol H3O{+} (i.e. H+) are required per mole CO3{2-}, 2(mol CO3{2-}) = (mol H3O{+} required to titrate) (.03725 L)(.1018 M) = 2(mol CO3{2-}) mol CO3{2-} = .001896 mol (.001896 mol CO3{2-} titrated) * (1 mol Na2CO3 / 1 mol CO3{2-}) * (106 g/mol) = .2010 g Na2CO3 (.2010 g Na2CO3 / .5376 g sample) * 100% = 37.38% Na2CO3 in sample

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.