Percent of Iron and Potassium in Iron Salt A 0.337 g of iron salt was put throug

Question

Percent of Iron and Potassium in Iron Salt A 0.337 g of iron salt was put through a cation exchange column and then titrated using 0.191 M NaoH as the titrant. It took 10.92 mt to reach the first equilvalence point and 20.23 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: Votume of titrant needed to titrate all of the potassium: 10.92 You are correct. your receipt no. is 161-1211 Previous Tries Mass of potassium found: 0.00999 Percentage of Potassium in salt: 372 Submit Answer Incoinect Tries 3/99 Previous Tries To solve for %iron: Volume of titrant needed to titrate all of the iron: 8.45 Tries 1/99 Previous Tries Submit Answer In Mass of iron found: Percentage of iron in salt: Submit Answer Tries 0/99 Preferences on what is marked as NEW iceadod View Chrondlogical view Sorting/Pittering aptions Export?

Explanation / Answer

Volume of NaOH required = 10.92 mL

number of moles of NaOH = 10.92 *10^-3 * 0.191 = 2.0857 * 10^-3 moles

so number of moles of POtassium = 2.0857 * 10^-3 moles

mass of potassium = 2.0857 * 10^-3 * 39.1 = 0.08155 g

percentage of potassium = 0.08155/0.337 * 100 = 24.2%

Volume of NaOH required = 20.23 - 10.92 = 9.31 mL

number of moles NaOH = 9.31 * 10^-3 * 0.191 = 1.78 * 10^-3 moles

so number of moles of Iron = 1.78 * 10^-3 moles

mass of Iron = 1.78 * 10^-3 * 56 = 0.09968 g

Percentage of Iron = 0.09968/0.337 * 100 = 29.58%

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