What concentration of nh4no3 is required to make [oh− ] = 1.0 × 10−5 in a 0.200-m solution of nh3?

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

We need a concentration of 0.361M for NH4NO3

Explanation:

Step 1: data given

[OH-] = 1.0 * 10^-5 M

Molarity of the NH3 solution = 0.200 M

Step 2: The balanced equation

NH3 + H2O --> NH4+ + OH-

Step 3: Calculate pOH

pH = pKa + log [A-]/[HA]

pOH = pKb + log [B+]/[BOH]

⇒ B+ = base's conjugate acid

⇒ BOH = the base

if [OH-] = 1.0 * 10^-5 M

then pOH = -log ( 1.0 * 10^-5)

pOH = 5

Step 4: Calculate [NH4+]

Kb of NH3 = 1.8 *10^-5

pKb = -log (1.8 * 10^-5) =  4.744

pOH = pKb + log [B+]/[BOH]

⇒pOH = 5

⇒pKb = 4.744

⇒[B+] = [NH4+] = TO BE DETERMINED

⇒[BOH] = [NH3] = 0.200 M

5 - 4.744 = log [NH4+]/[0.200]

0.256 = log [NH4+]/[0.200]

10^0.256 = [NH4+]/[0.200]

1.80 = [NH4+]/[0.200]

0.361 = [NH4+]

Step 5: Calculate [NH4NO3]

NH4NO3 --> NH4+ + NO3-

[NH4NO3] = 0.361 M

We need a concentration of 0.361M for NH4NO3

it is a solid

explanation:

5.6⋅10−^19  j

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