A 0.282 g of iron salt was put through a cation exchange column and then titrate

Question

A 0.282 g of iron salt was put through a cation exchange column and then titrated using 0.200 M NaOH as the titrant. It took 10.70 mL to reach the first equilvalence point and 18.77 mL to reach the second equilvalence point (volumes read directly off graph). What is the percent by mass of potassium and iron in the sample? To solve for % potassium: Volume of titrant needed to titrate all of the potassium: ????Tries 0/99 Mass of potassium found: Percentage of Potassium in salt: |????! Tries 0/99 To solve for %iron: Volume of titrant needed to titrate all of the iron: ????! Tries 0/99 Mass of iron found: Percentage of iron in salt:

Explanation / Answer

mass of sample = 0.282 g

Volume of titrant for all of potassium = 10.70 ml

moles NaOH used = 0.200 M x 10.70 ml = 2.14 mmol

moles potassium present = 2.14 mmol

mass potassium = 2.14 mmol x 39.1 g/mol = 0.084 g

percentage of potassium in the salt = 0.084 g x 100/0.282 g = 29.8%

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Volume of titrant used for all of iron = 18.77 ml

moles NaOH used = 0.200 M x 18.77 ml = 3.754 mmol

moles iron present = 3.754 mmol

mass iron = 3.754 mmol x 55.845 g/mol = 0.210 g

percentage of iron in the salt = 0.210 g x 100/0.282 g = 74.5%

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