Q18.181CP - Chemistry: Molecular Nature Of Matter And Change

Question

The molecular scene depicts the relative concentrations of ${\mathbf{\text{H}}}_{\mathbf{\text{3}}}{\mathbf{\text{O}}}^{\mathbf{\text{ + }}}$  (purple) and ${\mathbf{\text{OH}}}^{\mathbf{\text{ - }}}$ (green) in an aqueous solution at ${\mathbf{25}}^{\mathbf{0}}\mathit{C}$. (Counter ions and solvent molecules are omitted for clarity.) 

(a) Calculate the pH. 

(b) How many  ${\mathbf{\text{H}}}_{\mathbf{\text{3}}}{\mathbf{\text{O}}}^{\mathbf{\text{ + }}}$ ions would you have to draw for every ${\mathbf{\text{OH}}}^{\mathbf{\text{ - }}}$ ion to depict a solution of pH 4?

Step-by-Step Solution

Answer

a) $\text{pH = 7.5}$

b) Per ${\text{1H}}_{\text{3}}{\text{O}}^{\text{ + }}$ particle, there are $1\times {10}^6 O{H}^{-}$ particles in the solution.

Step 1: (a) To calculate pH

First, determine the $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\text{to}\left[{\text{OH}}^{\text{ - }}\right]$ ratio.

Purple $\left({\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right)\text{:2}$

Green $\left({\text{OH}}^{\text{ - }}\right)\text{:20}$

Let us divide the numbers with the smaller number, which is two, to get the ratio. Then designate an unknown variable to indicate that they are similar by a certain factor.

$\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right] \text{ = } \frac{\text{2}}{\text{2}}\text{x} \text{ = } \text{x}\left[{\text{OH}}^{\text{ - }}\right] \text{ = } \frac{\text{20}}{\text{2}}\text{x} \text{ = } \text{10x}$

Use the ${\text{K}}_{\text{w}}$ of water at ${25}^{0}C$ to solve for the unknown variable x.

$$\begin{gathered} {\text{K}}_{\text{w}}\text{ = } \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\left[{\text{OH}}^{\text{ - }}\right] \\ {\text{K}}_{\text{w}}\text{ = } \text{(x)(10x)} \\ {K}_w=10x^{2}x=\sqrt{\frac{K_w}{10}}=\sqrt{\frac{1\times {10}^{-14}}{10}} \\ x=3.16\times {10}^{-8} M \end{gathered}$$

Since $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\text{ = } \text{x} \text{ = } 3.16\times {10}^{-8}M$, then we can use this to solve for the pH.

$$\begin{gathered} \text{pH = } \text{ - log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right] \\ \text{=-log}\left(3.16\times {10}^{-8}\right) \\ \text{pH=7.5} \end{gathered}$$

Hence  $\text{pH = 7.5}$

Step 2: (b) To find the H 3 O  +  ions

First, solve for $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]$ .

$$\begin{gathered} \text{pH = } \text{ - log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right] \\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\text{ = } {\text{10}}^{\text{ - pH}} \\ \text{ = } {\text{10}}^{\text{ - 4}} \\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\text{ = } {\text{1×10}}^{\text{ - 4}} \end{gathered}$$

Next solve for $\left[{\text{OH}}^{\text{ - }}\right]$ using the ${\text{K}}_{\text{w}}$ of water at ${25}^0\text{C}$

$$\begin{gathered} {\text{K}}_{\text{w}}\text{ = }\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\left[{\text{OH}}^{\text{ - }}\right] \\ \left[{\text{OH}}^{\text{ - }}\right]\text{ = }\frac{{\text{K}}_{\text{w}}}{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]} \\ \text{ = }\frac{{\text{1×10}}^{\text{ - 14}}}{{\text{1×10}}^{\text{ - 4}}} \\ \left[{\text{OH}}^{\text{ - }}\right]{\text{ = 1×10}}^{\text{ - 10}} \end{gathered}$$

Now solve for the ratio by dividing $\text{the }\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\text{ by }\left[{\text{OH}}^{\text{ - }}\right]\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]}{\text{[OH - ]}}\text{ = }\frac{{\text{1×10}}^{\text{ - 4}}}{{\text{1×10}}^{\text{ - 10}}}{\text{ = 1×10}}^{\text{6}}$ Therefore, per ${\text{1H}}_{\text{3}}{\text{O}}^{\text{ + }}$ particle, there are  ${\text{1×10}}^{\text{6}}{\text{OH}}^{\text{ - }}$ particles in the solution.