Answer:
The concentration of NH₄NO₃ required to make [OH⁻] = 1.0 × 10⁻⁵ in a 0.200-m solution of NH₃ is 0.3639 M
Explanation:
Here we have
The reaction of Ammonia and water to produce OH⁻
NH₃ + H₂O → NH₄⁺ + OH⁻
From Henderson-Hasselbalch Equation for bases we have
pOH = pKb + log[acid]/[base]
The concentration of OH⁻ is given as [OH⁻] = 1.0 × 10⁻⁵
Therefore pOH = -log [OH⁻] = -log(1.0 × 10⁻⁵) = 5
The base dissociation constant is given by
kb = [B⁺][OH⁻]/[BOH]
For NH₃, kb = 1.8×10⁻⁵
pKb = -log Kb = 4.74
From the Henderson-Hasselbalch Equation
The acid concentration = [NH₄⁺] and the base concentration = [NH₃] = 0.200 M
pOH = pKb + log ([acid]/[base])
Substituting the values, we have
5 = 4.74 + log ([ NH4+ ]/0.200 M)
[NH4+] = 1.8197×0.200 M = 0.3639 M
NH₄NO₃ → NH₄⁺ + NO₃⁻
Therefore, since one mole of NH₄NO₃, produces one mole of NH₄⁺, the number of moles of NH₄NO₃ required to produce 0.3639 M of NH₄⁺ is 0.3639 M of NH₄NO₃
The concentration of NH₄NO₃ required to make [OH⁻] = 1.0 × 10⁻⁵ in a 0.200-m solution of NH₃ = 0.3639 M