Q167CP - Chemistry: Molecular Nature Of Matter And Change

Question

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 1000  atm (Kw = 1.101012). 

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

Step-by-Step Solution

Answer

$$\begin{gathered} \text{(a) }\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\text{ = 2.29}\times {\text{10}}^{\text{ - 7}} \\ \text{(b) }\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\text{ = 5.2}\times {\text{10}}^{\text{ - 12}} \\ \text{(c) }\left[{\text{OH}}^{\text{ - }}\right]\text{ = 5.2}\times {\text{10}}^{\text{ - 11}} \\ \text{(d) }\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\text{ = 1.10}\times {\text{10}}^{\text{ - 10}} \\ \text{(e) pH = 5.979} \end{gathered}$$

Step 1(a): Find [ H 3 O  +  ] in pure water

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

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

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

Step 2(b): Find [ H 3 O  +  ] in 0.010MNaOH

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

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.

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

Step 3: Find [ OH  -  ]  in 0.0010M HClO 4

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

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.

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

Step 4: Find [ H 3 O  +  ] in 0.0100M KOH

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

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

Step 5: Find the pH of pure water

(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.

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