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?

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 $\text{pH = 7.5}$

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 arerole="math" localid="1663312554471" ${\text{1×10}}^{\text{6}}{\text{OH}}^{\text{-}}$ particles in the solution.