🤓 Based on our data, we think this question is relevant for Professor Vasquez's class at OKSTATE.

We’re being asked to** calculate the pH** of a solution 0.201 mol NH_{4}CN and enough water to make 1.00 L of solution.

NH_{4}CN dissociates in water in the following manner:

**NH _{4}CN_{(s)} ⇌ NH_{4}^{+}_{ (aq)} + CN^{-} **

NH_{4}^{+}^{ }is the conjugate acid of a weak base, NH_{3}:

NH_{3} + H_{2}O ⇌ NH_{4}^{+} + OH^{-}; K_{b}=1.76x10^{-5}

CN^{-} is the conjugate base of a weak acid, HCN:

HCN + H_{2}O ⇌ CN^{-} + H_{3}O^{+ }; K_{a}= 4.9x10^{-10}

The pH of the salt of a weak acid and a weak base can be calculated using the following equation:

$\overline{){\mathbf{pH}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{(}{\mathbf{pK}}_{\mathbf{w}}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{pK}}_{\mathbf{a}}\mathbf{-}{\mathbf{pK}}_{\mathbf{b}}\mathbf{)}}$

*We will do the following steps to solve this problem:*

*Step 1: Calculate pK _{a} of the weak acid*

*Step 2: Calculate the pK _{b }of the weak base*

*Step 3: Determine pK _{w}*

*Step 4: Calculate the pH of the salt*

Calculate the pH of a solution prepared from 0.201 mol of NH_{4}CN and enough water to make 1.00 L of solution.