Q19.115CP - Chemistry: Molecular Nature Of Matter And Change

Question

It is possible to detect  NH₃ gas over ${\mathbf{\text{10}}}^{\mathbf{\text{ - 2}}}{\mathbf{\text{ M NH}}}_{\mathbf{\text{3}}}$. To what must be raised to form detectable ${\mathbf{\text{NH}}}_{\mathbf{\text{3}}}$ ?

Step-by-Step Solution

Answer

The deduction of  $\mathbf{0}\mathbf{.}\mathbf{01}\mathbf{\text{M}}$from $0.15\text{M}$  since that's the amount needed to make ${\text{NH}}_3$ .

Concept introduction

Ammonia, with the chemical formula NH3, is a colorless, pungent-smelling gas made up of nitrogen and hydrogen. Anhydrous ammonia is a highly pure form of ammonia that contains very little water. Under pressure, ammonia gas can be compressed into a liquid.

Calculation to form detectable NH 3

$$\begin{gathered} \mathrm{c}\left({\mathrm{NH}}_3\right)={10}^{-2} \mathrm{M} \\ \mathrm{c}\left({\mathrm{NH}}_4\mathrm{Cl}\right)=0.15 \mathrm{M} \end{gathered}$$

We only need to utilize the Henderson-Hasselbalch equation to solve this problem:

 $$\begin{gathered} \mathrm{Kw}={\mathrm{K}}_{\mathrm{a}}\times {\mathrm{K}}_{\mathrm{b}} \\ 1\times {10}^{-14}={\mathrm{K}}_{\mathrm{a}}\times 1.8\times {10}^{-5} \\ {\mathrm{K}}_{\mathrm{a}}=\frac{1\times {10}^{-14}}{1.8\times {10}^{-5}} \end{gathered}$$

  $$\begin{gathered} \mathrm{pH}=\mathrm{pKa}+\log \frac{\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{HA}\right]} \\ \mathrm{pH}= -\log \frac{1\times {10}^{-14}}{1.8\times {10}^{-5}}+\log \frac{1\times {10}^{-2}\mathrm{M}}{0.15\mathrm{M}-0.01\mathrm{M}} \end{gathered}$$

Therefore, we deduct $0.01\text{M}$ from  $0.15\text{M}$ since that's the amount needed to make  ${\text{NH}}_3$$\mathrm{pH}=8.11$.