Question

Suppose 0.829g of ammonium iodide is dissolved in 250.mL of a 30.0mM aqueous solution of potassium carbonate.

Calculate the final molarity of iodide anion in the solution. You can assume the volume of the solution doesn't change when the ammonium iodide is dissolved in it.

Round your answer to 3 significant digits.

Answer 1

K₂CO₃ (aq) + 2NH₄I (aq) ---- > 2KI(aq) + (NH₄)₂CO₃(aq)

Given that

0.829 g of ammonium iodide

Number of moles = amount in g / molar mass

= 0.829 g/ 144.94293 g/mol

= 0.00572 moles NH4I

One mole given one mole I-

So there are 0.00572 mole I- ions present.

In the reaction of potassium carbonate and ammonium iodide, potassium iodide and ammonium carbonate are formed which are ionic species and present as ion . \\ thus in the reaction mixture there are 0.00572 mole I- ions present.

Molarity = number of ions/ volume in L

= are 0.00572 mole I- ions /250*1/1000

= 0.25 M I- ions