Respuesta :
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)
[tex]0.26=log\frac{[NH_4^+]}{0.200 \hspace {0.09cm}M}[/tex]
[tex]10^{0.26} =1.8197=\frac{[NH_4^+]}{0.200 \hspace {0.09cm}M}[/tex]
[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
Answer:
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
