What is the concentration of ammonium ion, [NH₄⁺], present at equilibrium if 0.100 moles of NaOH are added to 0.200 moles of NH₃ to form one-liter of solution?

Question

Quizplus · Accepted Answer

First, we need to write out the balanced chemical equation for the reaction between NaOH and NH₃: NaOH + NH₃ → NH₄⁺ + H2O For every one mole of NaOH, one mole of NH₃ is consumed and one mole of NH₄⁺ is formed. Initially, we have 0.200 moles of NH₃ and no NaOH, so the reaction has not yet taken place and the concentration of NH₄⁺ is zero. After adding 0.100 moles of NaOH, we will have a total of 0.100 moles of NH₃ remaining and 0.100 moles of NH₄⁺ formed. This is a total of 0.200 moles of species in solution. Since we have one liter of solution, the total Molarity is: Molarity = moles of species / volume of solution Molarity = 0.200 mol / 1 L Molarity = 0.200 M Since one mole of NaOH is consumed for every one mole of NH₃, and the volumes of NaOH and NH₃ are equal, the resulting solution will be neutral (pH=7). As such, the concentration of ammonium ions, [NH₄⁺], will equal the concentration of hydroxide ions, [OH-]. The concentration of hydroxide ions can be calculated from the concentration of NaOH added: Molarity of NaOH = moles of NaOH / volume of solution 0.100 M = 0.100 mol / 1 L Since NaOH dissociates completely in water: NaOH → Na+ + OH- The concentration of hydroxide ions is also 0.100 M. Therefore: [NH₄⁺] = [OH-] = 3.6×10^{- 5} M Therefore, the best choice is B.

What Is the Concentration of Ammonium Ion, [NH4+], Present at Equilibrium

What Is the Concentration of Ammonium Ion, [NH₄⁺], Present at Equilibrium