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Explain the ionization of water and calculate the H3O+ concentration at 50°C.

Chemistry

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7.1

2.34×10−7 mol⋅dm−32.34 \times 10^{-7} \text{ mol}\cdot\text{dm}^{-3}

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Part 7.1· 3 steps

7.1.1. The ionization of water is ENDOTHERMIC.
As the temperature increases from 25 °C25 \text{ °C} to 50 °C50 \text{ °C}, the value of KwK_w increases from 1.00×10−141.00 \times 10^{-14} to 5.48×10−145.48 \times 10^{-14}. An increase in the equilibrium constant with increasing temperature indicates that the forward reaction (ionization) is favored by higher temperatures, which is characteristic of an endothermic process.

7.1.2.

  1. Identify the KwK_w value at 50 °C50 \text{ °C}.

    At 50 °C50 \text{ °C}, Kw=5.48×10−14K_w = 5.48 \times 10^{-14}.

  2. Use the relationship between KwK_w and [H3O+][\text{H}_3\text{O}^+] for pure water.

    For pure water, [H3O+]=[OH−][\text{H}_3\text{O}^+] = [\text{OH}^-].

    Kw=[H3O+][OH−]=[H3O+]2K_w = [\text{H}_3\text{O}^+][\text{OH}^-] = [\text{H}_3\text{O}^+]^2
  3. Calculate [H3O+][\text{H}_3\text{O}^+].

    [H3O+]=Kw[\text{H}_3\text{O}^+] = \sqrt{K_w} [H3O+]=5.48×10−14[\text{H}_3\text{O}^+] = \sqrt{5.48 \times 10^{-14}} [H3O+]=2.34×10−7 mol⋅dm−3[\text{H}_3\text{O}^+] = 2.34 \times 10^{-7} \text{ mol}\cdot\text{dm}^{-3}

    The concentration of H3O+\text{H}_3\text{O}^+ ions in water at 50 °C50 \text{ °C} is 2.34×10−7 mol⋅dm−3\boxed{2.34 \times 10^{-7} \text{ mol}\cdot\text{dm}^{-3}}.

    7.1.3. The pH of water will DECREASE.
    As the temperature increases, the value of KwK_w increases, which means the concentration of H3O+\text{H}_3\text{O}^+ ions increases. Since pH=−log⁡[H3O+]\text{pH} = -\log[\text{H}_3\text{O}^+], an increase in [H3O+][\text{H}_3\text{O}^+] leads to a decrease in pH.

Part 7.2

7.2.1. Water acts as an acid (donates a proton to HCO3−\text{HCO}_3^-):

H2O(l)+HCO3−(aq)⇌OH−(aq)+H2CO3(aq)\text{H}_2\text{O(l)} + \text{HCO}_3^-\text{(aq)} \rightleftharpoons \text{OH}^-\text{(aq)} + \text{H}_2\text{CO}_3\text{(aq)}

7.2.2. Water acts as a base (accepts a proton from HCO3−\text{HCO}_3^-):

H2O(l)+HCO3−(aq)⇌H3O+(aq)+CO32−(aq)\text{H}_2\text{O(l)} + \text{HCO}_3^-\text{(aq)} \rightleftharpoons \text{H}_3\text{O}^+\text{(aq)} + \text{CO}_3^{2-}\text{(aq)}

Part 7.3

7.3.1. An electrolyte is a substance that, when dissolved in a solvent (typically water), produces ions and is therefore capable of conducting electricity.

7.3.2. The most effective electrolyte is H2C2O4\text{H}_2\text{C}_2\text{O}_4.
The reason is that H2C2O4\text{H}_2\text{C}_2\text{O}_4 has the largest KaK_a value (6.0×10−26.0 \times 10^{-2}) among the given acids. A larger KaK_a value indicates a stronger acid, meaning it ionizes to a greater extent in solution, producing a higher concentration of ions and thus making it a more effective electrolyte.

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