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An Arrhenius base is a molecule that when dissolved in water will break down to yield an or hydroxide in solution.
An Arrhenius base is a compound that increases the OH− ion concentration in aqueous solution.
An Arrhenius base is a substance that, when dissolved in an aqueous solution, increases the concentration of hydroxide, or , ions in the solution.
Bases Properties
Arrhenius bases that are soluble in water can conduct electricity.
Bases often have a bitter taste and are found in foods less frequently than acids. Many bases, like soaps, are slippery to the touch.
Bases also change the colour of indicators. Red litmus turns blue in the presence of a base (see figure below), while phenolphthalein turns pink.
Some bases react with metals to produce hydrogen gas.
Acids (pH < 7.0) react with bases (pH > 7.0) to produce a salt and water. When equal moles of an acid and a base are combined, the acid is neutralized by the base. The resulting mixture will have a more neutral pH.
An Arrhenius acid is a substance that dissociates in water to form hydrogen ions or protons. In other words, it increases the number of H+ ions in the water. In contrast, an Arrhenius base dissociates in water to form hydroxide ions,
Example, sodium hydroxide, is added to an aqueous solution. NaOH dissociates into sodium, , and hydroxide,
ions.
Learn more about the Arrhenius bases here:
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Answer:
pH = 8.24
Explanation:
Nitrous acid (HNO₂) reacts with KOH, thus:
HNO₂ + KOH → KNO₂ + H₂O
Moles of HNO₂ are:
0.0257mL ₓ (0.370mol / L) = 0.00951moles.
In equivalence point, the complete moles of nitrous acid reacts with KOH producing potassium nitrite. There are needed:
0.00951mol ₓ (1L / 0.491mol) = 0.01937L ≡ 19.4mL of 0.491M KOH to reach equivalence point.
Total volume in equivalence point is: 19.4mL + 25.7mL = <em>45.1mL</em>
Potassium nitrite is in equilibrium with water, thus:
NO₂⁻ + H₂O ⇄ HNO₂ + OH⁻
Where equilibrium constant, Kb, is defined as:
Kb = 1.41x10⁻¹¹ =
In equilibrium, molarity of each compound are:
[NO₂⁻]: 0.00951mol/0.00451L - X = 0.211M - X
[HNO₂]: X
[OH⁻]: X
<em>Where X is reaction coordinate</em>
Replacing in Kb:
1.41x10⁻¹¹ =
0 = X² + 1.41x10⁻¹¹X - 2.97x10⁻¹²
Solving for X:
X = -1.72x10⁻⁶ <em>FALSE ANSWER. There is no negative concentrations.</em>
X = 1.72x10⁻⁶. <em>Right answer.</em>
That means:
[OH⁻]: 1.72x10⁻⁶M
As pOH is -log [OH⁻] and pH = 14-pOH:
pOH = 5.76; <em>pH = 8.24</em>