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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