Answer:
The molarity of the final ammonium cation is 0.252M
Explanation:
<u>Step 1:</u> Data given
Mass of ammonium chloride (NH4Cl) = 11.4 grams
Volume of 0.3 M aqueous solution of potassium carbonate (K2CO3) = 250 mL = 0.250L
<u>Step 2:</u> The balanced equation
2NH4Cl + K2CO3 → 2KCl + (NH4)2CO3
<u>Step 3:</u> Calculate moles of (NH4)Cl
moles (NH4)Cl = 11.4 grams /53.49 g/mol
Moles (NH4)Cl = 0.213 moles
<u>Step 4: </u>Calculate moles of K2CO3
Moles K2CO3 = Molarity * Volume
Moles K2CO3 = 0.3M * 0.250 L = 0.075 moles
<u>Step 5:</u> Calculate moles (NH4)Cl at the equilibrium
For 2 moles (NH4)Cl consumed, we need 1 mole of K2CO3 to produce 2 KCl and 1 mole of (NH4)2CO3
(NH4)2CO3l will dissolve in 2NH4+ + CO32-
Moles (NH4)2Cl = 0.213 moles - 2*0.075 = 0.063 moles
Moles NH4+ = moles (NH4)Cl = 0.063 moles
<u>Step 6:</u> Calculate Molarity of NH4+
Molarity = Moles / volume
Molarity of NH4+ = 0.063 moles / 0.250 L
Molarity of NH4+ = 0.252 M
The molarity of the final ammonium cation is 0.252M
