At $\mathbf{50}\mathbf{°}\mathbf{C}$and 1 atm,${\mathbf{\text{K}}}_{\mathbf{\text{w}}}\mathbf{\text{= 5.19}}\mathbf{\times }{\mathbf{\text{10}}}^{\mathbf{\text{- 14}}}$. At these conditions, find

(a)$\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$in pure water,

(b)$\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$in 0.010 MNaOH , and

(c) [OH] in${\mathbf{\text{0.0010MHClO}}}_{\mathbf{\text{4}}}$.

(d) Find$\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$in 0.0100 MKOH at$\mathbf{100}\mathbf{°}\mathbf{C}$and 1000atm (Kw = 1.101012).

(e) Find the pH of pure water at$\mathbf{100}\mathbf{°}\mathbf{C}$and 1000atm

$\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$in pure water

Note that ${\text{K}}_{\text{w}}\text{=}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]$ and pure water are neutral, so it will have

$\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}\left[{\text{OH}}^{\text{-}}\right]\text{. Solve for the}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]{\text{using theK}}_{\text{w}}\text{at}the \text{giv}en\text{conditions.}$

$\begin{array}{rcl}& & {\text{K}}_{\text{w}}\text{=}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]\\ & & {\text{K}}_{\text{w}}\text{=}{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}^{\text{2}}\\ & & \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}\sqrt{{\text{K}}_{\text{w}}}\\ & & \text{=}\sqrt{\text{5.19}\times {\text{10}}^{\text{- 14}}}\\ & & \text{= 2.29}\times {\text{10}}^{\text{- 7}}.\end{array}$

$\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{in 0.010MNaOH}$

First, write the equation for the dissociation of NaOH.

$\text{NaOH}\to {\text{Na}}^{+}{\text{OH}}^{-}$

$$\begin{gathered} \text{Since it is a strong base, its dissociation can be [NaOH] =}\left[{\text{Na}}^{\text{+}}\right]\text{=}\left[{\text{OH}}^{\text{-}}\right]\text{.} \\ \text{Therefore,}\left[{\text{OH}}^{\text{-}}\right]\text{= 0.010M}. \end{gathered}$$

Since we want to know the $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$we can solve it by using the $K_{w}$ at the given condition.

$\begin{array}{rcl}& & {\text{K}}_{\text{w}}\text{=}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]\\ & & \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{=}\frac{{\text{K}}_{\text{w}}}{\text{[OH - ]}}\\ & & \text{=}\frac{\text{5.19}\times {\text{10}}^{\text{- 14}}}{\text{0.010}}\\ & & \text{= 5.2}\times {\text{10}}^{\text{- 12}}.\end{array}$

$\left[{\text{OH}}^{\text{-}}\right]{\text{in 0.0010M HClO}}_{\text{4}}$

First, write the equation for the dissociation of${\text{HClO}}_4$.

${\text{HClO}}_4+{\text{H}}_2\text{O}\to {\text{H}}_3{\text{O}}^{+}{\text{ClO}}_4^{-}$

Since it is a strong acid,${\text{HClO}}_4+{\text{H}}_2\text{O}\to {\text{H}}_3{\text{O}}^{+}{\text{ClO}}_4^{-}$. Therefore $\left[{\text{H}}_3{\text{O}}^{+}\right]=0.0010\text{M.}$,Since we want to know the$\left[{\text{OH}}^{-}\right]$, we can solve it using the${\text{K}}_{\text{w}}$at the given condition.

$\begin{array}{rcl}& & {\text{K}}_{\text{w}}\text{=}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]\\ & & \left[{\text{OH}}^{\text{-}}\right]\text{=}\frac{{\text{K}}_{\text{w}}}{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}\\ & & \text{=}\frac{\text{5.19}\times {\text{10}}^{\text{- 14}}}{\text{0.0010}}\\ & & \text{=}\text{5.2}\times {\text{10}}^{\text{- 11}}.\end{array}$

$\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$in 0.0100M KOH

First, write the equation for the dissociation of KOH.

0.0100M KOH

Since it is a strong base,$\text{[KOH]}\text{=}\left[{\text{K}}^{\text{+}}\right]\text{=}\left[{\text{OH}}^{\text{-}}\right]$. Therefore_,.$\left[{\text{OH}}^{-}\right]=0.0100\text{M}$ Since we want to know the$\left[{\text{H}}_3{\text{O}}^{+}\right]$, we can solve it using the${\text{K}}_{\text{w}}$ at the given condition.

$\begin{array}{rcl}& & {\text{K}}_{\text{w}}\text{=}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]\\ & & \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\left.\text{=}\frac{{\text{K}}_{\text{w}}}{\text{[OH}}\right]\\ & & {\text{K}}_{\text{w}}\text{=}\frac{\text{1.10}\times {\text{10}}^{\text{- 12}}}{\text{0.0100}}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ & & \text{= 1.10}\times {\text{10}}^{\text{- 10}}.\end{array}$

(e) pH of pure water

Note that pure water is neutral, so it will have$\left[{\text{H}}_3{\text{O}}^{+}\right]=\left[{\text{OH}}^{-}\right]$. Solve for the$\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ using the${\text{K}}_{\text{w}}$ at the given condition.

$\begin{array}{rcl}& & {\text{K}}_{\text{w}}\text{=}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\left[{\text{OH}}^{\text{-}}\right]\\ & & \text{=}{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}^{\text{2}}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ & & \text{=}\sqrt{\text{1.10}\times {\text{10}}^{\text{- 12}}}\\ & & \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{= 1.05}\times {\text{10}}^{\text{- 6}}\end{array}$

Then solve for the pH.

$\begin{array}{rcl}& & \text{pH =}\text{- log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ & & \text{=}\text{- log}\left(\text{1.05}\times {\text{10}}^{\text{- 6}}\right)\\ & & \text{=}\text{5.979.}\end{array}$