To answer the given question step-by-step, let's analyze each part separately.
### Part (a): Determine if the autoionization of water is exothermic or endothermic.
We start by observing how the equilibrium constant for water's autoionization, [tex]\( K_w \)[/tex], changes with temperature. Here is the data provided:
\begin{tabular}{cc}
Temperature [tex]\(\left( {}^{\circ} \text{C} \right)\)[/tex] & [tex]\( K_w \)[/tex] \\
\hline
0 & [tex]\( 1.14 \times 10^{-15} \)[/tex] \\
25 & [tex]\( 1.00 \times 10^{-14} \)[/tex] \\
35 & [tex]\( 2.09 \times 10^{-14} \)[/tex] \\
40 & [tex]\( 2.92 \times 10^{-14} \)[/tex] \\
50 & [tex]\( 5.47 \times 10^{-14} \)[/tex] \\
\end{tabular}
From the data, we can see that as the temperature increases, the value of [tex]\( K_w \)[/tex] also increases.
Concept in Chemistry:
If the equilibrium constant for a reaction increases with temperature, it generally indicates that the reaction is endothermic. This is because, for an endothermic reaction, increasing the temperature shifts the equilibrium to favor the formation of products, which increases the value of the equilibrium constant.
Conclusion:
The autoionization of water is endothermic.
### Part (b): Calculate [tex]\(\left[ \text{H}^+ \right]\)[/tex] and [tex]\(\left[ \text{OH}^- \right]\)[/tex] in a neutral solution at [tex]\(50^\circ \text{C}\)[/tex].
Given:
- [tex]\( T = 50^\circ \text{C} \)[/tex]
- [tex]\( K_w \text{ at } 50^\circ \text{C} = 5.47 \times 10^{-14} \)[/tex]
We know that in a neutral solution, the concentrations of hydrogen ions [tex]\(\left[ \text{H}^+ \right]\)[/tex] and hydroxide ions [tex]\(\left[ \text{OH}^- \right]\)[/tex] are equal:
[tex]\[ \left[ \text{H}^+ \right] = \left[ \text{OH}^- \right] \][/tex]
Let [tex]\( x \)[/tex] be the common concentration of [tex]\(\left[ \text{H}^+ \right]\)[/tex] and [tex]\(\left[ \text{OH}^- \right]\)[/tex]. Therefore,
[tex]\[ x \cdot x = K_w \][/tex]
[tex]\[ x^2 = 5.47 \times 10^{-14} \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ x = \sqrt{5.47 \times 10^{-14}} \][/tex]
From the calculation, we find:
[tex]\[ x = 2.3388031127053 \times 10^{-7} \][/tex]
Therefore:
[tex]\[ \left[ \text{H}^+ \right] = 2.3388031127053 \times 10^{-7} \, \text{M} \][/tex]
[tex]\[ \left[ \text{OH}^- \right] = 2.3388031127053 \times 10^{-7} \, \text{M} \][/tex]
Conclusion:
At [tex]\( 50^\circ \text{C} \)[/tex], in a neutral solution:
- The concentration of hydrogen ions [tex]\(\left[ \text{H}^+ \right]\)[/tex] is [tex]\( 2.3388031127053 \times 10^{-7} \, \text{M} \)[/tex].
- The concentration of hydroxide ions [tex]\(\left[ \text{OH}^- \right]\)[/tex] is [tex]\( 2.3388031127053 \times 10^{-7} \, \text{M} \)[/tex].
