**We’re being asked to c alculate the ratio [NH_{3}]/[NH_{4}^{+}] in ammonia/ammonium chloride buffered solutions for ph 9.00. **

In an **ammonia/ammonium chloride buffered solutio**n, ammonia, **NH _{3}**

Recall that for buffers the** pH can be determined** using the Henderson-Hasselbalch equation:

$\overline{){\mathbf{pH}}{\mathbf{=}}{{\mathbf{pK}}}_{{\mathbf{a}}}{\mathbf{+}}{\mathbf{log}}\left(\frac{\mathbf{}\mathbf{weak}\mathbf{}\mathbf{base}}{\mathbf{conjugate}\mathbf{}\mathbf{acid}}\right)}$

where:

[NH_{3}] = concentration of the **weak ****base, NH _{3}**

[NH_{4}^{+}] = concentration of **conjugate ****acid, NH _{4}^{+ }**

Since **we're given the pH, **we can solve for the ratio: [NH_{3}]/[NH_{4}^{+}]

But first, we need to determine K_{a}, Since ammonia is a weak base:

**K _{b} = 1.76 x 10**

Recall that we can **s****olve K**_{a }from **K**_{b} using the equation:

$\overline{){{\mathbf{K}}}_{{\mathbf{w}}}{\mathbf{=}}{{\mathbf{K}}}_{{\mathbf{a}}}{\mathbf{\times}}{{\mathbf{K}}}_{{\mathbf{b}}}}\phantom{\rule{0ex}{0ex}}\overline{){{\mathbf{K}}}_{{\mathbf{a}}}{\mathbf{=}}\frac{{\mathbf{K}}_{\mathbf{w}}}{{\mathbf{K}}_{\mathbf{b}}}}$

***K _{w} = 1.0 x 10*

Calculate the ratio [NH_{3}] / [NH_{4}^{+}] in ammonia/ammonium chloride buffered solutions with the following pH values:

a. pH = 9.00

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