Q176CP - Chemistry: Molecular Nature Of Matter And Change

Question

(a) If ${\mathbf{\text{K}}}_{\mathbf{\text{w}}}\mathbf{\text{ = 1.139}}\mathbf{\times }{\mathbf{\text{10}}}^{\mathbf{\text{ - 15}}}$at $\mathbf{0}\mathbf{°}\mathbf{C}$ and $\mathbf{\text{5.474}}\mathbf{\times }{\mathbf{\text{10}}}^{\mathbf{\text{ - 14}}}$at $\mathbf{50}\mathbf{°}\mathbf{C}$, find and pH of water at $\mathbf{0}\mathbf{°}\mathbf{C}$and $\mathbf{50}\mathbf{°}\mathbf{C}$. 

(b) The auto ionization constant for heavy water (deuterium oxide, ${\mathbf{\text{D}}}_{\mathbf{\text{2}}}\mathbf{\text{O}}$) is $\mathbf{\text{3.64}}\mathbf{\times }{\mathbf{\text{10}}}^{\mathbf{-}\mathbf{16}}$ at $\mathbf{0}\mathbf{°}\mathbf{C}$and $\mathbf{\text{7.89}}\mathbf{\times }{\mathbf{\text{10}}}^{\mathbf{\text{ - 15}}}$at $\mathbf{50}\mathbf{°}\mathbf{C}$ . Find $\left[{\text{D}}_{\text{3}}{\text{O}}^{\text{ + }}\right]$and pD of heavy water at $\mathbf{0}\mathbf{°}\mathbf{C}$and $\mathbf{50}\mathbf{°}\mathbf{C}$. 

(c) Suggest a reason for these differences.

Step-by-Step Solution

Answer

a) $$\begin{gathered} \mathrm{At} 0°\mathrm{C}, \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\text{ = } \text{3.375}\times {\text{10}}^{\text{ - 8}} \text{M and pH = 7.4717.} \\ \mathrm{At} 50°\mathrm{C}, \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\text{ = } \text{2.340}\times {\text{10}}^{\text{ - 7}}\text{M and} \text{pH = 6.6308.} \end{gathered}$$

b) $$\begin{gathered} \mathrm{At} 0°\mathrm{C}, \left[{\text{D}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\text{ = 1.91}\times {\text{10}}^{\text{ - 8}}\text{ and Mpp = 7.719.} \\ \mathrm{At} 50°\mathrm{C}, \left[{\text{D}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\text{ = 8.88}\times {\text{10}}^{\text{ - 8}}\text{ and MpD = 7.051.} \end{gathered}$$

c) At higher temperatures, the reaction will shift to the right and will dissociate more, resulting in a lower pH.

Step 1(a): Find the [ H 3 O  +  ] and the pH of water at 0 ° C and 50 ° C .

Given information:

At $\text{0}°{\text{C, K}}_{\text{w}}\text{ = 1.139}\times {\text{10}}^{\text{ - 15}}$

At $\text{50}°{\text{C, K}}_{\text{w}}\text{ = 5.474}\times {\text{10}}^{\text{ - 14}}$

At . $\text{0}°\text{C, solve the }\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\text{ and the pH}$Note that water is neutral, therefore, its$\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\text{ and }\left[{\text{OH}}^{\text{ - }}\right]$  are equal.

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

$\begin{array}{rcl}& & {\text{K}}_{\text{w}}\text{ = } \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\left[{\text{OH}}^{\text{ - }}\right]\\ & & \text{ = } {\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]}^{\text{2}}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\text{ = } \sqrt{{\text{K}}_{\text{w}}}\\ & & \text{ = } \sqrt{\text{1.139}\times {\text{10}}^{\text{ - 15}}}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\\ & & \text{ = } \text{3.375}\times {\text{10}}^{\text{ - 8}}\text{M.}\end{array}$

$\begin{array}{rcl}& & \text{pH = } \text{ - log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\\ & & \text{ = } \text{ - log}\left(\text{3.375}\times {\text{10}}^{\text{ - 8}}\right)\\ & & \text{ = } \text{7.4717.}\end{array}$

At . $\text{50}°\text{C, solve the }\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\text{ and the pH}$Note that water is neutral, therefore, its are $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\text{ and }\left[{\text{OH}}^{\text{ - }}\right]$equal.

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

$\begin{array}{rcl}& & {\text{K}}_{\text{w}}\text{ = }\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\left[{\text{OH}}^{\text{ - }}\right]\\ & & \text{ = }{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]}^{\text{2}}.\end{array}$

$$\begin{gathered} \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\text{ = }\sqrt{{\text{K}}_{\text{w}}} \\ \text{ = } \sqrt{\text{5.474}\times \text{ - 14}} \\ \text{ = } \text{2.340}\times {\text{10}}^{\text{ - 7}}\text{M.} \\ \text{pH = } \text{ - log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right] \\ \text{ = } \text{ - log}\left(\text{2.340}\times {\text{10}}^{\text{ - 7}}\right) \\ \text{ = } \text{6.6308.} \end{gathered}$$

Therefore:

$$\begin{gathered} \text{At} \text{0}°\text{C, }\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right] \text{ = } \text{3.375}\times {\text{10}}^{\text{ - 8}}\text{M} \text{and pH = 7.4717.} \\ \text{At} \text{50}°\text{C, }\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right] \text{ = } \text{2.340}\times {\text{10}}^{\text{ - 7}}\text{M} \text{and} \text{pH = 6.6308.} \end{gathered}$$

Step 2(b): Find [ D 3 O  +  ] and pD of heavy water at 0 ° C and 50 ° C

Given information:

At $\text{0}°\text{C, K = 3.64}\times {\text{10}}^{\text{ - 16}}$

At $\text{50}°\text{C, K = 7.89}\times {\text{10}}^{\text{ - 15}}$

At $\text{0}°\text{C, solve the }\left[{\text{D}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\text{ and the pD}$. Note that water is neutral, therefore, its are $\left[{\text{D}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\text{and}\left[{\text{OD}}^{\text{ - }}\right]$ equal.

$\left[{\text{D}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\text{ = }\left[{\text{OD}}^{\text{ - }}\right]$

$\begin{array}{rcl}& & \text{K = }\left[{\text{D}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\left[{\text{OD}}^{\text{ - }}\right]\\ & & \text{ = }{\left[{\text{D}}_{\text{3}}{\text{O}}^{\text{ + }}\right]}^{\text{2}}.\end{array}$

At $\text{0}°\text{C, solve the }\left[{\text{D}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\text{ and the pD}$. Note that water is neutral, therefore, its $\left[{\text{D}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\text{and }\left[{\text{OD}}^{\text{ - }}\right]$ are equal.

$\left[{\text{D}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\text{ = }\left[{\text{OD}}^{\text{ - }}\right]$

$\begin{array}{rcl}& & \text{K = } \left[{\text{D}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\left[{\text{OD}}^{\text{ - }}\right]\\ & & \text{ = } {\left[{\text{D}}_{\text{3}}{\text{O}}^{\text{ + }}\right]}^{\text{2}}\\ & & \\ & & \left[{\text{D}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\text{ = } \sqrt{\text{K}}\\ & & \text{ K} \text{ = }\sqrt{\text{3.64}\times {\text{10}}^{\text{ - 16}}}\\ & & \left[{\text{D}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\text{ = 1.91}\times {\text{10}}^{\text{ - 8}}\text{M}\\ & & \text{ pD = } \text{ - log}\left[{\text{D}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\\ & & \text{ = } \text{ - log}\left(\text{1.91}\times {\text{10}}^{\text{ - 8}}\right)\\ & & \text{ = 7.719.}\end{array}$

At $\text{50}°\text{C, solve the }\left[{\text{D}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\text{ and the pD}$. Note that water is neutral, therefore, its $\left[{\text{D}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\text{and }\left[{\text{OD}}^{\text{ - }}\right]$ are equal.

$\left[{\text{D}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\text{ = }\left[{\text{OD}}^{\text{ - }}\right]$

 $\begin{array}{rcl}& & \text{K = }\left[{\text{D}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\left[{\text{OD}}^{\text{ - }}\right]\\ & & \text{ = }{\left[{\text{D}}_{\text{3}}{\text{O}}^{\text{ + }}\right]}^{\text{2}}\\ & & \left[{\text{D}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\text{ = }\sqrt{\text{K}}\\ & & \text{ K} \text{ = }\sqrt{\text{7.89}\times {\text{10}}^{\text{ - 15}}}\\ & & \left[{\text{D}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\text{ = 8.88}\times {\text{10}}^{\text{ - 8}}\text{M}\\ & & \text{ pD = } \text{ - log}\left[{\text{D}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\\ & & \text{ = } \text{ - log}\left(\text{8.88}\times {\text{10}}^{\text{ - 8}}\right)\\ & & \text{ = } \text{7.051.}\end{array}$

 

 

Therefore:

$$\begin{gathered} \text{At 0}°\text{C, }\left[{\text{D}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\text{ = 1.91}\times {\text{10}}^{\text{ - 8}}\text{M}\mid \text{pp = 7.719.} \\ \text{At 50}°\text{C, }\left[{\text{D}}_{\text{3}}{\text{O}}^{\text{ + }}\right]\text{ = 8.88}\times {\text{10}}^{\text{ - 8}}\text{M}\mid \text{pD = 7.051.} \end{gathered}$$

Step 3(c): Suggest a reason for the differences

As observed in the two set of examples (a and b), at low temperatures, the neutral pH is higher. At higher temperatures, the neutral pH is lower. The auto ionization of water is endothermic and at higher temperatures, the reaction shifts to the right and will dissociate more. More $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}\right]$  will be produced, resulting in a lower 

Therefore, at higher temperatures, the reaction will shift to the right and will dissociate more, resulting in a lower pH.