Question

In this lab exercise a solution containing K+ is passed down an ion exchange column containing a cation exchange resin in the acid form. The K+ ions in the solution replace the resin bound H3O+ ions which elute from the column and then are titrated. This means that the column now holds the K+ ions. To use the column again we must "recharge" the column to remove K+ ions. In your Lab Notebook, describe the exact procedure you will follow to "recharge" the exchange column for the next use. A 0.173 gram solution of KxFe(C2O4)yzH2O is dissolved in water and the solution is passed down a column containing a cation exchange resin in the acid form. The eluted solution is then titrated with a standard NaOH solution to the first equivalence point. Experimentally it required 9.66 mL of a 0.152 M NaOH solution to reach the first equivalence point.

Calculate the number of moles of K⁺ in the weighted sample.

Calculate the mass (grams) of K⁺ in the weighted sample

Calculate percent of potassium in the weighted sample.

Answer 1

Part A

Procedure for recharge the column

The used up cation exchange resin is regenerated by treatment with moderately concentrated sulfuric acid or HCl.

K⁺-resin) + HCl (aq)  $\small \rightarrow$ KCl + H₃O⁺- resin

then washed the column with water.

Part B

at neutralization point

moles of acid = moles of base

moles of H₃O⁺ = moles of OH^-

In the given problem , at equivalence point

[ mole = molarity * volume (L) ] given volume of NaOH = 9.66 ml = 9.66*10^-3 L

moles of NaOH = moles OH^- = 0.152 *9.66*10^-3 = 0.00147 = moles of H₃O⁺  

now ,

Moles of K⁺ bound = moles of H₃O⁺ eluted = 0.00147

hence moles of K⁺ in the sample = 0.00147 moles

now, mass of K⁺ in the weighted sample = moles of K⁺ * atomic mass of K = 0.00147*39 = 0.0572 g

percentage of potassium in the weighted sample = (mass of K^+**100) / mass of the sample

= (0.0572 *100)/0.173 = 33.10 %