Hydroxide Ion Concentration Of Strong Acids Chemistry Tutorial
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Calculating the Hydroxide Ion Concentration of Strong Monoprotic Acids
Before the acid is added to water, water molecules are in equilibrium with hydrogen ions and hydroxide ions:
| water | | hydrogenions | + | hydroxide ions |
| H2O | | H+(aq) | + | OH-(aq) |
Only a very small number of water molecules dissociate into H+(aq) and OH-(aq). At 25°C, [H+(aq)] = [OH-(aq)] ≈ 10-7 mol L-1 (a very low concentration) and Kw = [H+(aq)][OH-(aq)] = 10-7 × 10-7 = 10-14
When a strong monoprotic acid is added to water, the acid dissociates completely to form H+(aq) and an aqueous anion:
| acid | → | hydrogenions | + | anions |
| HA | → | H+(aq) | + | A-(aq) |
Adding the acid to the water disturbs the water dissociation equilibrium: H2O
H+(aq) + OH-(aq) By Le Chatelier's Principle, adding more H+(aq) to the water will shift the equilibrium position to the left. The water dissociation equilibrium system responds to the addition of more H+(aq) by reacting some of the H+(aq) with some of the OH-(aq) in order to re-establish equilibrium. So, increasing the concentration of H+(aq) in the water, reduces the concentration of OH-(aq), but, the water dissociation constant does not change4, Kw is still 10-14. So, Kw = [H+(aq)][OH-(aq)] = 10-14
| At 25°C | [H+(aq)]mol L-1 | [OH-(aq)]mol L-1 | Kw |
|---|---|---|---|
| pure water(before acid added) | 10-7 | 10-7 | 10-14 |
| aqueous solution(after acid added) | > 10-7 | < 10-7 | 10-14 |
We can use the value of Kw and [H+(aq)] to calculate [OH-(aq)] at a given temperature: At 25°C: Kw = [H+(aq)][OH-(aq)] = 10-14 By rearranging this equation (formula) we can determine the concentration of hydroxide ions in the aqueous solution: [OH-(aq)] = 10-14 ÷ [H+(aq)] Both [H+(aq)] and [OH-(aq)] must be in units of mol L-1 (mol/L or M)
If we know the pH of the acidic solution, we know the concentration of hydrogen ions in the solution because pH = -log10[H+(aq)] By reaarranging this equation (formula) we can find the concentration of hydrogen ions in mol L-1: [H+(aq)] = 10-pH This value of [H+(aq)] can then be used to calculate [OH-(aq)] in mol L-1 using the appropriate value for Kw.
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